Sign In | Subscribe
INSTRUCTORS Raffi Hovasapian John Zhu
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Calculus AB
  • Discussion

  • Download Lecture Slides

  • Table of Contents

  • Transcription

Bookmark and Share
Lecture Comments (2)

1 answer

Last reply by: Professor Hovasapian
Fri Apr 1, 2016 3:05 AM

Post by Kaushik Srinivasan on March 27 at 11:36:25 PM


In your example 9 (IX) , I got a=0.25, b=0, c=-3 and d=4.

This gives the equation 0.25x^3 - 3x + 4. This has tangents of gradient 0 (making horizontal tangents) at (-2,8) and (2,0)

I even graphed it on Desmos and it worked. Did I do something wrong and accidentally graph the derivative?

Differentiation of Polynomials & Exponential Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Differentiation of Polynomials & Exponential Functions 0:15
    • Derivative of a Function
    • Derivative of a Constant
    • Power Rule
    • If C is a Constant
    • Sum Rule
    • Exponential Functions
  • Example I: Differentiate 7:45
  • Example II: Differentiate 12:38
  • Example III: Differentiate 15:13
  • Example IV: Differentiate 16:20
  • Example V: Differentiate 19:19
  • Example VI: Find the Equation of the Tangent Line to a Function at a Given Point 12:18
  • Example VII: Find the First & Second Derivatives 25:59
  • Example VIII 27:47
    • Part A: Find the Velocity & Acceleration Functions as Functions of t
    • Part B: Find the Acceleration after 3 Seconds
    • Part C: Find the Acceleration when the Velocity is 0
    • Part D: Graph the Position, Velocity, & Acceleration Graphs
  • Example IX: Find a Cubic Function Whose Graph has Horizontal Tangents 34:53
  • Example X: Find a Point on a Graph 42:31

Transcription: Differentiation of Polynomials & Exponential Functions

Hello and welcome back to, and welcome back to AP calculus.0000

Today, we are going to start talking about ways of taking the derivative of a function, of different types of functions but doing it in a quick way.0005

Let us just jump right on in.0013

We have seen and we have been dealing with the definition of the derivatives.0016

I will go stick with blue today.0022

We have seen that the derivative of any function can be gotten from the definition of the derivative.0029

That is we take the limit as h approaches 0 of f(x) + h - f(x) divided by h.0048

We formed this difference quotient, we simplified it as much as possible.0061

The last thing that we do is we take h to 0 to see what happens.0064

We end up getting this new function.0069

We need a quicker way to find derivative.0072

We cannot just keep using the definition of the derivative.0075

You saw from previous lessons that it can get very tedious.0078

We need a quicker way, you need a quicker way to find derivatives.0082

In this particular lesson, we are going to lay out the basic rules for finding derivatives of polynomial functions and exponential functions.0101

For the majority of these lessons, I’m actually not going to be proving where we get these formulas0112

because our main concern is having the formulas at our disposal, being able to use them.0116

We want to develop technique.0121

For those of you that go onto higher mathematics, you actually revisit this stuff again 0123

and you will spend time actually proving all of these things.0127

If you want to take a look at the proofs, they are in your textbook.0131

They are very straightforward, it is actually no different than anything that you have done so far.0135

You are just using the definition of a derivative in its most general way.0139

And then, you run through it, you prove it, they are very easy to follow.0144

But again, it just tends to be a little bit tedious. We would not worry about proofs.0147

The first thing we are going to talk about is the derivative of a constant, let us start there.0153

The derivative of a constant.0162

If you guys do not know already from previous work, the derivative with respect to the variable of a constant C is just equal to 0.0166

The derivative of 5 is 0, the derivative of E is 0, the derivative of 9.7 is 0.0179

Number 2, we will just call it the power rule.0188

This says that if n is any real number, it can be an integer, it could be a fraction, it could be radical to, whatever, any real number.0208

Then, ddx of x ⁺n is equal to n × x ⁺n-1.0225

In other words, we take the n, we bring it out front, turn it into a coefficient.0235

And then, we change the exponent, we drop it by 1.0240

If the exponent is 5, the exponent becomes 4.0244

If the exponent is 1/3, the exponent, the n-1 becomes 1/3 -3/3 becomes -2/3, that is it.0246

This is the basic rule for all polynomials.0255

Let us see, next rule, if C is a constant then the derivative of C × some function f(x), whatever it happens to be.0261

Basically, it just says that the derivative of a constant × a function is the constant × the derivative of a function.0284

We can pull out the constant and then just save it and multiply it by everything, after we take the derivative of a function, equal C × ddx of whatever f happens to be.0290

I’m going to leave out the x just to make it a little bit more notationally tractable.0303

Let us do the sum rule.0321

If f and g are differentiable, then the derivative of f + g is just equal to,0329

I’m going to leave it as it is.0356

The derivative of f + the derivative of g.0362

In other words, if I have some sum of a function, I can just take the derivative individually and add them up, that is it.0368

The derivative of the sum is the sum of the derivatives, that is all.0374

You can think of it as distributing the derivative operator over what the sum is.0378

We can differentiate term by term and add.0386

The last is the exponential functions.0389

The exponential functions, we have the derivative of e ⁺x = e ⁺x.0400

The exponential function is very special.0409

When you take the derivative of it, you get the function back.0414

That is very special, it is going to play a very huge role in all the mathematics that we do.0417

This is going to be ddx of a ⁺x.0423

In this case, we have just e, 2.718, it is this.0429

Whenever there is a different constant, whatever it is, some a ⁺x, the derivative of that is,0434

the a ⁺x stays but we have to multiply it by the nat-log of the base a.0442

These are our basic rules for finding the derivatives of polynomial and the exponential functions.0452

With that, let us just launch right into our examples because that is when everything starts to make sense.0460

The first example, 9x⁵ – 4x³ + 14x² – x – 12.0467

Pretty standard polynomial, 5th degree in this case.0476

We have a sum, this is one polynomial but it is made up of 12345 terms.0480

We are going to use the sum rule.0487

We are just going to differentiate one term at a time.0488

A combination of a power rule and the sum rule.0492

The power rule says, take this coefficient, bring it down here, and then subtract by 1.0496

Because of this 9 in front of it, it is a constant, we can pull out and save it for later.0503

This is actually going to be 9 × 5x⁴ -4 ×, the 3 comes down, 4 × 3x².0507

Subtract the exponent by 1, + 14, I bring the 2 down, I put here and this becomes just x¹.0522

Same thing here, I bring the 1 over here and this is going to be x¹ -1 is 0, x⁰ is just 1.0534

Basically, any time you are taking the derivative of 1x, 2x, 3x, 4x, it is just the number itself, the coefficient.0544

The derivative of a constant is 0.0551

Of course, we simplify it.0554

We have 45x⁴ -12x², 14 × 2x + 28x – 1.0556

There you go, that is the derivative of that polynomial, nice, straightforward, very easy.0572

An application of the power rule, an application of the constant multiplied by a function, and application of the sum rule.0579

When we do these, we are sort of doing them.0588

We can say yes, we are applying this rule or that rule.0592

After you do four of these, you are expert at it already.0596

Let us look at a portion of the graph just for this particular one.0605

I just wanted to take a look at it.0609

What I’m going to do is we are going to go ahead and take a look at what the graph of that looks like, 0611

and then, the graph of the derivative on the same graph.0615

Let us see, there we go.0622

We have this one right here which is our f(x) that was the original function.0626

And then, this right here, this is actually the derivative.0634

This is f’(x).0637

We have seen this before, we have dealt with it before,0643

when we dealt with graphs and taking a look at the function and the derivative on the same graph.0645

Notice, here the slope of this original function is positive, one is positive.0652

It is positive but as it comes up here, the slope is actually decreasing.0660

The line from your perspective, the slope is actually going like this, it is becoming 0.0665

The slope is positive, this is the derivative.0671

That is what the derivative is, it is a slope, it is positive.0674

When it hits the top of the graph, it hits 0.0677

Of course now the slope of this curve is negative.0680

It is negative, it goes down.0683

At some point, it starts to turn and starts to turn and become 0 again.0686

As this point rises and it hits 0, and then from that point on, the function is rising, the slope is rising.0692

The line is here, the slope is increasing and becoming positive.0704

That is why you have this.0711

The red is the original function, the blue is the derivative.0714

It is kind of interesting, even after all these years, whenever I look at a function and its derivative on the same graph, 0721

it always takes me a minute to look at it and make sure I'm concentrating on the right thing.0726

When you are looking at the original function, you are looking at the derivative.0734

What the derivative is describing is the behavior of the slope on the original function.0739

This line, that is the tangent one that is becoming that way and going down that way, and this way, and this way.0743

That is what you are doing, that is what the derivative represents.0753

Let us go ahead and differentiate this function, √x - 4√x.0761

This radical symbol is a leftover from years and years ago, hundreds of years ago.0768

I do not care for the symbol myself but again it is ubiquitous in mathematics.0776

We deal with this just like turning everything into fractional exponents.0780

I’m going to rewrite this f(x) as x¹/2 – x¹/4.0786

I can go ahead and apply the power rule to this.0796

Very simple.0799

Again, this comes down, f’(x) is going to equal ½ x and ½ -1 because we subtract 1 from the exponent.0801

This becomes - ½.0812

This, the ¼ comes down, this becomes ¼ x.0815

And then, ¼ - 1 is - ¾. 0820

That is it, that is your derivative.0824

You are absolutely welcome to leave it like that.0826

It is perfectly good mathematical symbolism.0829

You can write it as follows, if you want to.0833

This is really going to have to do with your teacher and what it is that they want.0851

You should ask them, is it okay to leave it in this form, do you write in radical form?0855

One of the things you are going to notice, if you have not noticed it, you are going to notice it now that 0861

we actually start do taking the derivatives of functions very quickly, 0864

especially when we get to product and quotient rule in the next lesson.0867

There are several degrees to which you can simplify a function.0875

At some point, you are just going to have to stop.0878

You are going to have to ask your teacher, where it is okay to stop.0880

If you want, you can write this as of 1/ 2 √x.0884

This x⁻¹/2, bring it down to the denominator, and then put the radical sign back on there.0890

-1/4, this is ¾, this is going to be the 4√x³.0895

This is another way to write it, if you want it to.0905

Example number 3, f(t) = (t/4)³.0912

Let us go ahead and expand this out.0920

We have f of T = t³/ 4³.0922

4 × 4 is 16, 4 × 16 is 64.0930

This is nothing more than 1/64 t³.0933

We can go ahead and treat it, this is the constant, this is the exponent.0937

Therefore, f’(t) is equal to 1/64 × 3t².0942

We end up with 3/64 t² or 3t²/ 64.0952

However it is that you want to write it out.0962

I have always like the constant to be separate but that is just my own personal taste.0966

Example number 4, x³ + 3x² + 2x + 3/ √x.0975

This one again, in the next lesson we are going to be doing something called the quotient rule.0983

In this particular case, we notice that there is only one thing in the dominator.0990

We can actually put each of these terms over that dominator and see if it simplifies, which it does.0995

I can rewrite f of x as x³ / x¹/2, because √x is x ^ ½ + 3x²/ x¹/2 + 2x/ x¹/2 + 3/ x¹/2.1003

This becomes 3 - ½, this is now the same base x, I can just subtract the exponents.1028

I get 3 - ½ is 2 ½, this is going to be x⁵/2 + 3x - ½ is 1 ½, 3x³/2.1038

2x¹/2, 1 – ½ + 3x⁻¹/2.1055

Now I have x to all these different coefficients.1063

I just differentiate, f’(x) is equal to 5/2x.1067

5/2 – 1 is 5/2 – 2/2 which is going to be 3/2 + 3 × 3/2x.1076

3/2 – 1 is ½, ½ + 2 × ½, I bring this down.1088

I subtract 1 from that ½ -1 is -½.1100

And then of course, we have 3 × -1/2 x – 1/2 -1 is -3/2.1105

My final answer f'(x) is equal to 5/2x³/2 + 9/2x ^½.1120

The 2 and the 2 cancel.1133

We are left with +x ^-½.1136

This is -3/2x⁻³/2.1141

There you go, that is your f’(x).1147

I hope that made sense.1155

F (s) is s² – 2/3√s⁴.1160

We are getting pretty accustomed to dealing with this now.1165

This is just s² -2.1167

I will do s, this 4/3, and when I bring it up, it is going to be - 4/3.1172

Is that correct, yes it is.1185

We have f(s), I got to tell you that it is easy to make mistakes in differentiation.1189

One of the things about calculus is it is not particularly difficult in terms of the application of the technique.1198

It is just sort of keeping track of all the little things.1204

In calculus, it is the details that matter, the individual little details, a - sign.1207

Remembering to subtract one from the exponent.1213

Remembering to add one to the exponent, coefficients, all of these things to keep track of.1216

I can guarantee you, I’m going to be making my fair share of mistakes on this.1221

Vigilance, that is all we can try to do is remain vigilant.1225

This is going to be s⁴/3, when we bring it up -4/3.1231

When we take the derivatives of this, this is going to be 2s -2 × -4/3s.1236

Now, -4/3 -1 is - 7/3.1249

F’ (s) = 2s, - and - becomes a +.1257

2 × 4 is 8, we have 8/3s⁻⁷/3.1264

There you go, this is our derivative.1272

Find the equation of a tangent line to the following function at the given point.1281

Graph both the function and the tangent line on the same screen.1286

We have our function f(x) = x² + √x.1290

We want to find the equation of a line that is the tangent line.1296

In other words, that is the derivative.1302

I’m sorry, no, the equation of the tangent line through this point.1305

The slope of that line is going to have is the derivative of this function at that point.1311

Because that is what the derivative is, the derivative is the slope of the line of the tangent line to the graph at that point.1318

That was one of our interpretations of the derivative.1327

The derivative is a rate of change.1330

It is the rate at which y is changing, when I make a small change in x.1331

It is also the slope of the tangent line to the graph at that particular x and y value, at that particular point.1336

Let us go ahead and find the derivative first.1345

We have f(x), let us rewrite it as x² + x¹/2.1350

We have f’(x) which is going to be 2x + ½ x⁻¹/2.1358

In this particular case, because I’m going to be putting values in, I’m just going to go ahead and write it in a form that is more familiar.1370

+1/2 × √x.1378

This is the formula for the derivative.1385

When we put this x value into there, we actually get the slope of the line.1388

We want to find f’(2).1394

That is going to equal 2 × 2 + ½ √2.1399

When I work that out, I get 4.35355.1407

This is our slope, it is the slope of the tangent line.1415

When I put the x value into the derivative function that I get, it gives me the slope of the tangent line at that point.1423

I go ahead and find the line itself.1430

The line is going to be 1 -y1 = the slope m × x - x1.1433

I have x and I have y, that is the point that it passes through.1441

I get y -5.414 = 4.35355 × x-2.1446

I’m going to go ahead and leave it in this form.1460

It is up to your teacher if they want you to actually multiply this out, simplify fractions that might show up.1462

However is it that they want to see it.1468

Perhaps, they want to see it in ax + by= c form, personal choice.1470

That is it, nice and straightforward.1476

Let us go ahead and take a look at what this actually looks like.1480

This right here, this is f(x), that is the function.1486

That point right there is our 2 and 5.414, that is the scales, the scales are not the same for the x and y axis.1491

I sort of expanded them out.1502

This is the equation of the tangent line.1503

This is the tangent line, the equation that we got.1507

This is the equation for the tangent line.1512

This is not f’(x).1520

F’(x), f’, in this particular, is f’(2).1523

It is the slope of the tangent line.1531

The derivative is the slope.1533

The derivative is not the line itself, very important.1535

The slope changes, if we hit peak to different point, it is going to be a different slope.1539

If we hit to peak a point up here, it is going to be a different slope.1544

Keep repeating it over and over again.1548

The derivative is the slope of the tangent line not the equation of the tangent line.1552

Find both the first and the second derivatives of e ⁺x -∛x.1561

Very simple, we are just going to take the derivative twice.1566

Let me rewrite f(x).1570

F(x) = e ⁺x -, I will write this as x¹/3.1580

We have f’(x) =, the derivative of the exponential function is the function itself.1588

That is just e ⁺x – 1/3x⁻²/3.1595

Bring this down, 1/3 – 1, the exponent – 1 becomes -2/3.1602

That is f’, now I will go ahead and do f”(x).1608

We take the derivative of the derivative, the first derivative, this becomes the second derivative.1614

The derivative of the exponential function is e ⁺x.1619

This is -1/3, now we take the derivative of this.1623

We bring this down, x-2/3 – 1- 2/3- 3/3 is - 5/3.1629

Our f’(x) and another notation for that is d ⁺2y dx², we have seen that notation before, =e ⁺x.1641

- and - is +, 2/9x⁻⁵/3.1653

Nice and straightforward.1663

S = 3t³-6t² + 4t + 2 is the equation of motion of a certain particle, with s in meters and t in seconds.1672

This is the position function.1681

This is the position function, in other words at any time t, let us say t is 5 seconds, I put T in there.1686

What I will end up getting is where the particle is, along the x axis tells me where it is.1692

We want you to find the following.1702

Find the velocity and the acceleration functions as functions of time,1704

the acceleration after 3 seconds, the acceleration when the velocity is 0.1707

Graph the position of the velocity and acceleration graphs on the same screen.1714

Let us go ahead and find the velocity and acceleration functions as functions of time.1721

Again, we have dealt with these before.1725

Whenever you get in the position function, the velocity function is the first derivative.1727

The acceleration function is the second derivative.1731

Velocity = ds dt or s’, however you want to do it, that is equal to, 3 × 3 is 9.1736

We dropped the exponent by 1, 9t².1748

2 × 6, the 2 comes down, 2 × 6 is 12.1753

It becomes -12t and +4 because 1 × 4 is 4 and t¹-1 is t0.1758

It ends up going away.1768

This is the velocity function.1769

The acceleration that = the derivative of the velocity with respect to t, that is equal to the second derivative of the position function d² s dt² .1772

Now I just take the derivative of the velocity.1785

2 × 9 is 18, that is 18t-, I take the derivative of this.1788

There we go, this is my velocity function, this is my acceleration function.1793

At any time t, I just plug it in to get the velocity and the acceleration.1798

At t = 2, the position is this, it is going this fast and it is accelerating with that acceleration.1803

Nice and straightforward.1810

Part B, they want you to find the acceleration after 3 seconds.1814

That is nice and easy, they just want the acceleration at 3.1820

We use this, 18 × 3 – 12.1824

I hope to God that I did my arithmetic correctly.1829

I’m notorious for making arithmetic mistakes but I think the answer should be 42 m/s².1831

That is the unit of acceleration.1838

Velocity is in meters per second.1841

The position is in meters, velocity is in meters per second.1843

Acceleration is in meters per second which is m/s².1845

Part C, they say they want the acceleration when the velocity is 0.1852

We have to set the velocity function to 0.1865

Find the values of t at which the velocity is 0, then plug those t values back in the acceleration equation.1868

Let us do exactly what it says.1874

We set the V equal to 0.1878

We have to find t, then use the acceleration function.1884

The velocity as a function of t is equal to 9t² -12t + 4.1893

I set that equal to 0 and I solve for t.1903

In this particular case, I’m not going to go through the particulars of this quadratic equation.1907

I hope that is not a problem.1911

I'm presuming that you have your calculator at your disposal or perhaps you just want to do it by hand,1913

by completing the square or using the quadratic formula.1918

In this particular case, I do not think it could be factored, whatever it is that you need to do by all means.1921

T ends up being 0.6667, one answer.1927

When a quadratic equation has only one answer not two, that means it just touches the graph, that touches the x axis.1935

Our acceleration that we are going to be looking for is the acceleration 0.6667.1945

That is equal to 18 × 0.6667 -12.1952

Our acceleration ends up being 0.1959

Let us go ahead and take a look at the graph of all three on the same screen, on the same page.1964

Here is our s(t), this is our position function that was the cubic equation.1973

This one right here, the blue, this is our S‘, this is our velocity function v(t).1980

This one right here, I think it is purple or magenta, this is S", this is the acceleration function.1990

Notice, positive slope, velocity is positive.1999

The velocity is positive, it is above the x axis.2009

The particle is moving to the right.2011

The position, the derivative actually goes to 0.2018

The slope goes to 0 that is why it hits that and it becomes positive again, and it increases.2028

The derivative follows the slope of the tangent line along that.2035

The acceleration, the acceleration is the derivative of the derivative.2042

Now, this parabola that we have, the velocity function is the function.2046

The derivative of that is going to be a straight line.2052

The slope is negative but it is increasing, negative but it is increasing.2058

At this point, the slope hits 0 and then the slope become positive.2064

That is why you have from negative to positive.2070

The function, its first derivative, the velocity function, its second derivative, the acceleration.2074

Nice and straightforward, I hope.2084

Let us go ahead and try this one.2091

A little bit longer, a little bit more involved.2095

Again, it is not like you have a lot at your disposal.2097

What you have at your disposal right now is this idea of the derivatives, we go with that.2101

Find the cubic function, ax³ + bx³ + cx + d, whose graph has horizontal tangents at -2,8 and 2,0.2107

In this particular case, we need to find a, we need to find b, we need to find c, and we need to find d, four variables that we need to find.2118

Let us go ahead and write this out.2128

F(x) = ax³ + bx² + cx + d.2130

We have 4 variables.2142

We know really only one way to find 4 variables, when we are given this little information.2145

We need to find 4 equations.2150

We know that if we have 4 equations and 4 unknowns, theoretically, we can find all of the unknowns.2152

In this case, a, b, c, and d.2157

We are going to be looking for 4 equations.2159

Where those equations are going to come from.2161

We know f(-2), we have it right here f(-2) is 8.2164

We just put -2 in for x and we put 8 equal to 8.2175

When I put -2 in for here, we get -8a + 4b, I’m putting -2 in for x, + 4b -2c + d = 8.2184

That is my first equation.2205

We also know f(2), f(2)=0.2207

Again, I can put 2 in for x and get something at a, b, c, and d, that is equal to 0.2210

We know f(2), f(2) that becomes 8a + 4b + 2c + d = 0.2220

This is our second equation, we are doing well.2232

They also tell us something else.2237

They tell us that the graph actually has horizontal tangents at these points.2238

Horizontal tangent means that the graph itself, no matter what the graph looks like, still the slope is 0.2243

Therefore, the slope is the derivative.2253

I take the derivative of this thing to get a new equation and I set that equal to 0 at these points.2256

I plug in -2 and 2 into the derivative of this to get two other equations.2266

Let us write all that out.2272

Horizontal tangents at x = -2 and x =2, that means a slope of 0.2276

This implies that f’ of x at 2 and -2 = 0.2297

A slope of 0 means the derivative is 0.2311

Let us take f’(x).2314

F’(x) is equal to 3ax² + 2bx + c.2316

The derivative of d is 0.2331

Now let us find, let us form the equation f’(-2).2334

F’(-2), -2 we are putting into the x value.2339

-2² is 4, 4 × 3 is 12.2344

We have 12a.2348

-2 and 2 this is -4b + c.2350

We know that the slope is actually equal to 0.2355

We have our third equation.2360

Now let us do f’ at 2.2362

F’ at 2 is going to be 12a + 4b + c.2364

That slope is also equal to 0.2371

There we have it.2374

We have our first equation, we have our second equation, we have our third equation, and we have our fourth equation.2376

4 variables, 4 equations, 4 equations, 4 unknowns.2384

Let us go ahead and work out the rest here.2391

We have 4 equations and 4 unknowns.2393

Our unknowns are a, b, c, and d.2396

Let us rewrite the equations again.2401

We have -8a + 4b -2c + d, that = 8.2405

That is fine, I guess I can do it over here.2420

We have the equation 8a + 4b + 2c + d = 0.2423

We have 12a -4b + c = 0.2435

We have 12a + 4b + c = 0.2441

Again, I’m not going to go through the process of actually solving these equations simultaneously.2448

It just has to do with, you can enter it into your software.2452

You can do it on your calculator.2456

You can do it at any number of online free equation solvers, that is how I did it.2458

Or you can do it by hand, I also did it by hand just to double check.2466

By fiddling around with some equations, solving for one variable here, solving for the same variable here.2469

Putting into these equations, whatever technique that you are going to use.2475

Again, these are techniques from previous mathematics.2479

We are just using them in support of what it is that we are doing now, with just calculus.2482

However you solve this, when you do, you end up with a = 0.1667.2487

You end up with b = 0.2496

You end up with c = -2.2498

You end up with d = 5.333.2501

Our f(x) equation, the equation that we wanted is 0.1667 x³ - 2x is +0x², I will just leave that out, +5.333.2507

There you go, nice and straight forward.2533

Again, we have the functions, we have the points, the x and y values, 2535

and they say horizontal tangent so we know that the derivative is 0.2543

We can create 4 equations in those 4 unknowns.2546

Let us see, find the points on the graph of the function f(x) = 1 + 3 e ⁺x-2x 2554

with tangent line at such a point is parallel to the line, the 2x – y = 4.2562

We have f(x) that is equal to 1 + 3 e ⁺x – 2x.2570

Let us go ahead and take the derivative of this, point at the tangent line2579

because the slope of the tangent line is going to be the derivative at that particular point.2586

F'(x) = 0, it is going to be 3 e ⁺x-2.2591

They say that the tangent line, the tangent line at such a point is parallel to the line 2x – y = 4.2600

Over here, I have the line 2x - y = 4.2609

This is y = 2x-4.2615

If the tangent line is going to be parallel to this line, it is going to have the same slope.2622

It is going to have a slope of 2.2627

The f'(x) is an equation for the slope.2633

Therefore, I’m going to set this equal to 2.2637

F'(x), we want f'(x), the slope of the tangent line to equal 2 because it is parallel.2643

Let us form that equation.2655

Let me do it over here.2661

We have 3 e ⁺x-2 = 2.2663

We get 3 e ⁺x = 4/3.2672

We take the nat-log of both sides.2678

We get x = nat-log of 4/3.2681

When we do this, the nat-log of 4/3 is 0.2877, that is the x value.2691

We take this x value.2700

They want the point on the graph of the function where the tangent line at such a point is parallel to the line.2703

We found the x value of that point.2709

If we want the y value, we have to put it back into the original function.2712

We get f(0.2877) that is equal to 4.247.2717

Our point is 0.2877, 4.247, there we go.2732

Once again, find the point on the graph of the function f(x) where the tangent line at such a point is parallel to the line.2749

I took the derivative of this function.2755

The derivative of that function at a given point is going to be a slope of the tangent line through that point.2758

They say that the line, the tangent line is parallel to this.2767

The slope of this line is 2.2775

I set the derivative equal to 2 to find the x value.2777

I found this x value, I plugged it back in the original equation to find the y value.2780

It passes through this point and has a slope of 2.2786

Thank you so much for joining us here at

We will see you next time, bye. 2793