WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we are going to be discussing operations on functions, beginning with arithmetic operations.
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Recall that two functions can be added, subtracted, multiplied, or divided.
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The domain of the sum, difference, product, or quotient is the intersections of the domains of the two functions.
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And remember that intersection, when you are talking about sets, is the areas of the sets that overlap.
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Often, we will be talking about domains that are all real numbers, and so then there would be complete overlap of those two.
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But if the sets are slightly different (the domains are slightly different), the domain of these is only
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the intersection of the domains of the two functions you are performing the operation on.
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We will talk about when we are working with division, because that is a special case, and there are some additional restrictions on the domain.
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But let's just go ahead and start out with addition.
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Let's take, as an example, the two functions f(x) = 2x² + 5, and a second function g(x) = 3x² - 4.
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OK, if you were to add these, then the notation would be as follows: f + g(x).
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And this can be rewritten to show that it really means f(x) + g(x).
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So, f + g(x) is just the two functions added together.
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Looking at what my two functions are: I have 2x² + 5 as the first function,
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and I am going to be adding that to my second function, which is 3x² - 4.
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And recall that, when you are adding polynomials, all you need to do for addition is remove the parentheses.
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And that, when we go ahead and add like terms, is going to give us 5x².
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And then, combining the constants, 5 - 4 is 1.
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So, the sum of these two functions is 5x² + 1.
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OK, now talking about subtraction: if we were to subtract f - g(x), this can be considered as f(x) - g(x).
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So, it is the same idea as addition, but you just have to be more careful with the signs.
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So here, we are going to be taking (2x² + 5) - (3x² - 4).
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Now, when you remove the parentheses, for the first one--this first expression here--this is really positive in front of it;
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so removing the parentheses is simple--you just take them away; the signs stay the same.
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To remove the parentheses for this second expression, you have to reverse the signs.
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So, when I apply the negative sign to 3x², that is going to give -3x².
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Applying the negative sign to the -4 will give me positive 4.
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And then, I combine these: I get like terms, 2x² - 3x², to get -x²; and 4 + 5 (5 + 4) gives me 9.
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So, that was addition and subtraction of two functions; now, let's talk about multiplication.
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f times g(x) is the same as f(x) times g(x); this is the same as multiplying two binomials, as we have done in earlier work in this course.
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So, it is (2x² + 5) times (3x² - 4); and since these are binomials,
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I use the FOIL method: multiplying the first two terms gives me 4 times 3 is 6x⁴;
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my outer terms are going to give me -8x²; the inner two terms...5 times 3...that is 15x².
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And finally, 5 times -4 gets -20.
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Simplify by combining like terms: I combine the two x² terms, and that is going to give me...
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-8x² + 15x², is going to give me positive 7x²; minus 20.
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OK, so this is multiplication of two functions.
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Finally, division: division has that additional restriction on the domain, as I mentioned.
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Let's look at dividing here: f divided by g(x), which is actually the same as f(x) over g(x).
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f(x) is 2x² + 5...over 3x² - 4.
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Now, when you are handling division, you need to think about excluded values in the denominator.
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As always, if the denominator is 0, that would result in an expression that is undefined.
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You cannot divide by 0, so we have to think about situations where this expression,
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3x² - 4, would equal 0, and then exclude those from the domain.
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So, I am going to go ahead and solve for x, and see what values of x would make this expression 0.
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I can do that by adding 4 to both sides, dividing both sides by 3, and then taking the square root of both sides.
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This tells me that the excluded values, when performing this operation (performing division and finding the quotient) are x = √ or -√4/3.
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So, x cannot equal ±√4/3, because if it does, the result will be a 0 in the denominator and a value that is undefined.
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So, this is a situation that applies only to division.
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Composition of functions: suppose that f and g are functions, and the range of g, the second function, is a subset of the domain of f, the first function.
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I will talk a little bit more about that in a second.
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But right now, let's look at what the notation is, and what it is really saying when we talk about composition of functions.
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The composition of f and g is defined by...this symbol, this open circle, means "composed with."
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It is not a multiplication symbol: the open circle would be read as "f composed with g of x."
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And that is going to result in a composite function.
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And another way to write this, and to think about it, is that what this is saying is f(g(x)).
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Something to keep in mind as we work with composition of functions is that you actually move from right to left when you are working with these.
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So, when I consider a composite function (and it could actually be composed of three functions or four functions),
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I am going to start out by working my way from the right, seeing if I am given a value here--
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instead of g(x), it could be g(2)--and then working with that, determining what g(x), and then plugging that value into my f function.
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OK, so let's first use an example: if I have two functions--one is f(x) = x² - 2x + 1,
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and another one is g(x) = 4x - 5--I might be asked to find the composition of those two functions.
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And that would be...I could be asked to find f composed with g, or g composed with f; but I am going to work with f composed with g.
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Again, I am rewriting that as...this is saying it is f of g of x.
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So, I am going to start here with this function on the right.
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And I am going to say, "OK, this is really saying f(g(x))."
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Well, I look here, and g(x) is 4x - 5.
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Previously, we have talked about evaluating a function for an algebraic expression.
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We first started out by evaluating functions for numbers, like f(3): I would just substitute all of the x's with 3's.
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I also talked about finding f for an algebraic expression; and that is really what we end up doing here.
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So, if I am trying to find f(4x - 5), wherever I see an x, I just replace it with this expression, 4x - 5...
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which is going to be...here I have x², so I am going to replace that x with 4x - 5 and square it.
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I have a second x right here; I have -2 times whatever is in here, since that is replacing x.
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And then, I am adding 1; so right here, where there is an x, simply substitute 4x - 5.
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And then, we are going to figure what this would become equal to.
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And this is going to give me...if you recall, if I square a binomial, if I form a perfect square trinomial from that...
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it ends up being this first term squared (4x, squared), and then 2 times the product of these two terms;
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so this is going to give me 4x, times -5, plus the square of this last term.
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You could always write this out as (4x - 5) (4x - 5) and use FOIL if you didn't remember that rule.
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OK, then over here, I am going to multiply everything inside the parentheses by -2 to get -8x + 10 + 1.
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OK, so this is going to give me...I am actually rewriting this like that, because I also have to square the 4.
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So, this is going to give me (4x)², so that will actually be 16x²,
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and then 2 times 4x is 8x, times -5 is going to give me 8x times -5...-40x.
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And then here, I have -5 times -5, so that is + 25, minus 8x plus 10 plus 1.
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Simplifying, I still have 15x²; combining -40x and -8x is going to be -48x.
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25 + 10 + 1 is 35...36.
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f composed with g: I went about finding this by first replacing this with g(x), which is 4x - 5,
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and then finding f of that by replacing all of the x's here with 4x - 5.
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And then, I squared this binomial, took 2 times the binomial here, and then added 1.
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And this is the result: f composed with g of x equals this.
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Now, talking a little bit more about domain and range: let's say that we were asked to find f composed with g(2).
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Well, I know what f composed with g(x) is, and I could just plug the 2 in here.
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But let's say we hadn't done that work--that we were just given these two, and then told, "OK, find f composed with g(2)."
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So, I am just going to start there and think about how this works.
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Well, I start at the right; I am going to rewrite this here, first, as f(g(x)).
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In this case, I said that x is going to be 2; so I am going to put a 2 there.
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So, I am going to find g(2), and then I am going to find f of that value.
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So, this is going to equal f of...well, g(x) is 4x - 5, but I am asked to find g(2).
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So remember that g(2) is going to be 4 times 2, minus 5.
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I am going to put that in here: 4 times 2, minus 5.
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OK, that equals f of 8 - 5, which is 3, so I am being asked to find f(3).
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Now, let's think about this for a second: here, my result, 3, is an element of the range of g.
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My input value for g--my domain value--is 2; I evaluated that for this function g, and I came up with 3.
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g(2) is 3, so 3 is part of the range.
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Now, I am using that as my input value for the function f.
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So, 3 is an element of the range of g, and an element of the domain of f.
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The range of g is a subset of the domain of f; that is what this is saying up here.
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So, I evaluate g for a particular value--just the general case x, or a certain number.
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I find what that is: that is a part of this range; it is also an element of the domain of f.
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The entire set of values--the entire range for g--is a subset of the domain of f.
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So, let's go ahead and finish this out: I found g(2) to be 3; now I am going to evaluate f for 3.
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Well, f is x² - 2x + 1, so wherever I see an x, I am simply going to substitute in a 3.
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That is going to give me 9 - (6 + 1); that is 9 - 7, so that is 2.
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So, f composed with g(2)...actually, that is 4--a correction right there: f composed with g(2) is 4.
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So again, starting out, we are going right to left, finding g(2), substituting what I came up with,
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which is 3, right in here, finding f(3), and evaluating for that.
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And I find that f composed with g(2) is 4.
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Up here, I just talked about the general case, f composed with g(x), and I found this.
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OK, so that is an introduction to composition of functions.
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And one important thing to keep in mind is that composition is not commutative.
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What we mean by that is that, in general, f composed with g does not equal g composed with f.
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We will see, later on, cases where we will talk about this more, later;
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but in general, you can't just flip these two around and assume that the opposite case is equal, because it is actually usually not.
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Illustrating that with an example: let's let f(x) equals x² + 1, and g(x) equal 2x - 3.
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Now, I am going to find both f composed with g and g composed with f, and then compare what I get.
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So, f composed with g(x) = f(g(x)); starting from the right, let's put g(x) in here, which is 2x - 3.
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Now, I am evaluating f for this expression, 2x - 3.
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And f is x² + 1, so I am going to substitute 2x - 3 for this x, and square it, and then add 1.
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Squaring this would give me 2x squared, plus 2 times 2x times -3, plus -3 squared, plus 1.
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OK, that gives me 4x²; 2 times 2 is 4, times -3 is -12x; plus -3², which is 9, plus 1.
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Simplify to 4x² - 12x + 10.
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f composed with g(x) is equal to this; now, let's try g composed with f(x), which is equal to g(f(x)).
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f(x) is x² + 1; I am going to put that in here.
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Now, I am going to evaluate g for x² + 1 by substituting x² + 1 everywhere there is an x in this function.
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That is going to give me 2 times x² + 1 - 3.
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2 times x² is 2x²; plus 2 times 1 is 2; minus 3 gives me 2x² - 1.
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And this is g composed with f(x).
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As you can see, f composed with g(x) is not the same as g composed with f.
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So, I can't make that assumption, that these two are the same.
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In fact, if anything, it is likely that they are not the same; composition is not commutative.
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The first example: in this example, we are given two functions, f(x) and g(x), and asked to find their sum and their difference.
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Starting out with addition: recall that f + g(x) = f(x) + g(x).
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So, I am simply going to add the two functions; f(x) is 3x³ - 2x² - x - 1,
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plus -4x⁴ + 2x³ - 3x - 4.
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OK, since this is addition, I can simply remove the parentheses without worrying about having a problem with the signs.
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The signs remain the same, so I just continue to retain the signs in here, since this is addition.
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I can just take the parentheses away.
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Let's add this, and at the same time put it in descending order to help keep track of everything.
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The largest power I have is 4; so I have -4x⁴, and there is no like term that I can combine that with.
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For terms that are cubed, I have 3x³ and 2x³ to give me 5x³.
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For terms that are squared, I only have one term, and that is -2x².
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For x's, I have -x and -3x to give -4x; and for constants, -1 and -4 are -5.
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So, f + g(x) equals this expression.
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Subtraction: f - g(x) = f(x) - g(x); now here, I do need to be careful with the signs.
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So, I am rewriting: 3x³ - 2x² - x - 1, minus g(x), which is 4x⁴ + 2x³ - 3x - 4.
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OK, removing the parentheses: since this is a positive in front of this, it remains 3x³...
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Here, I need to apply the negative sign to each term inside the parentheses; and I can do that by reversing the sign.
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So, this becomes -4x⁴; -2x³; a negative and a negative--that gives me a positive 3x; and a negative and a negative...plus 4.
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Combining like terms, I get -4x⁴; here, I have 3x³ and -2x³, leaving + x³.
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I only have one x² term, so I leave that alone; that is -2x².
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For x's, -x + 3x is going to leave me with just 2x, and then the constants: -1 + 4 gives me _ 3.
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So, f - g(x) is given by this expression right here.
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Example 2: this time, we are given two functions, f(x) and g(x), and told to find the product and the quotient of these functions.
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OK, beginning with multiplication: the product f times g(x) equals f(x) times g(x),
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which equals (4x² - 7) times (3x² + 9x).
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So, the first terms--multiplying those, that is 12x⁴; the outer terms give me 4 times 9; that is 36x² times x, so that is 36x³.
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The inner two terms--that is -21x²; and then -7 times 9 is going to give me -63; and I have an x here.
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OK, and looking at this, I can't simplify this any further, because I don't have any like terms.
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So, I am just going to leave that alone; and this gives me f times g of x.
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OK, the second task is to find the quotient, f divided by g of x, which equals f(x) divided by g(x).
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which equals (4x² - 7) divided by (3x² + 9x).
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Now, recall: when working with division of functions, we need to find the excluded values.
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And the excluded values are those values of x, those numbers of the domain, that will make the denominator 0,
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because if this denominator is 0, I will have an undefined expression.
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So, any value of x that makes 3x² + 9x equal 0 is excluded from the domain.
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Handling this by factoring: I can see that I have a greatest common factor of 3x.
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I have a 3 here and a 3 in here, and I have an x here and an x in here that I can pull out.
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This will leave behind an x and a 3.
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Using the zero product property, this tells me that 0 = 3x, or it could be that x + 3 = 0.
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If either of those is true, this product becomes 0.
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OK, so this will end up giving me (let's rewrite this as) 3x = 0, so divide both sides by 3: x = 0--that is an excluded value.
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And x = -3--these are excluded values, meaning this function is not defined for domain values of x = 0 and x = -3.
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Those values are excluded from the domain.
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Example 3: now, we are going to work with composition of functions.
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Given f(x) and g(x), we are asked to find f composed with g(x) and g composed with f(x), starting with this one.
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OK, so recall that f composed with g(x) is the same as saying f(g(x)).
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So, I am going to start from the right and go to the left.
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First, I am going to figure out what g(x) is; and I have that given right here as 4x².
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I am asked to find f of that; I am going to evaluate f for 4x².
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So, wherever I see an x in f(x), in this function, I will substitute 4x².
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So, this is going to give me 2 times x; and in this case, x is 4x²; minus 3.
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What I have here is rewritten here, just with 4x² replacing the x.
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OK, that is going to give me 8x² - 3; so, f composed with g of x equals 8x² - 3.
00:29:13.000 --> 00:29:26.300
Now, g composed with f(x) equals g(f(x)), starting from the right: what do I have in here?
00:29:26.300 --> 00:29:34.500
I have f(x), which is 2x - 3; so this equals g(2x - 3).
00:29:34.500 --> 00:29:44.800
This means I need to evaluate the function g when x is defined as 2x - 3.
00:29:44.800 --> 00:29:55.200
So, I have an x here; it is 4 times x², so 4(2x - 3)², just replacing this with this.
00:29:55.200 --> 00:30:16.600
OK, squaring 2x - 3 and multiplying that times 4 is going to give me (2x)² + 2 times 2x times -3 + (-3)².
00:30:16.600 --> 00:30:25.800
That is going to give me 4 times...this is 2 times 2...that is going to give me 4x²;
00:30:25.800 --> 00:30:33.500
2 times 2x is 4x, times -3; that is -12x; (-3)² is 9.
00:30:33.500 --> 00:30:46.400
Multiplying each of these by 4 gives me 16x²; 4 times -12 is -48x; 4 times 9 is 36.
00:30:46.400 --> 00:30:54.200
And this also illustrates what we talked about before, that composition is not commutative.
00:30:54.200 --> 00:31:04.000
f composed with g of x is not equal to g composed with f of x.
00:31:04.000 --> 00:31:09.900
It may be, but it is not necessarily true; you cannot assume that it is true.
00:31:09.900 --> 00:31:16.300
This time, we are going to be working with three functions: f(x), g(x), and h(x).
00:31:16.300 --> 00:31:23.700
And we handle these the same way as we did composition of functions, when we were only working with two functions.
00:31:23.700 --> 00:31:26.900
And we are going to work from right to left.
00:31:26.900 --> 00:31:37.200
Here, instead of just asking for f or g or h of x, they are asking me for f(g(h(the particular value -2))).
00:31:37.200 --> 00:31:46.300
OK, so I am going to start out by looking for h(-2).
00:31:46.300 --> 00:31:53.200
Let's go ahead and find that value; I want to find f(g(h(-2))).
00:31:53.200 --> 00:32:06.300
So, what is h(-2)? Well, h is 2x² + 3; therefore, h(-2)...I would have to substitute in 2...I put a -2 here and square it...+ 3.
00:32:06.300 --> 00:32:32.000
So, this is f(g(2 times -2 squared, plus 3)), which equals f(g(...-2 squared is 4, plus 3)).
00:32:32.000 --> 00:32:51.500
So, this gives me 8 + 3, which is f(g(11)); so starting from the right, I evaluated h for -2; that gave me 11.
00:32:51.500 --> 00:33:01.600
Now, this is a member of the range of h, and it is also an element of the domain of g, because I am plugging it in here as an input value.
00:33:01.600 --> 00:33:21.900
So now, I need to find g(11): well, g(x) equals 4x - 2; so substituting 4 times 11 - 2 is what I am going to be doing here.
00:33:21.900 --> 00:33:40.600
f(4 times 11 minus 2) equals 44 - 2, equals 42; so, g(11) = 42.
00:33:40.600 --> 00:33:51.100
Finally, I am just left with f(42); so I need to evaluate f(x) when x is 42.
00:33:51.100 --> 00:34:03.000
So, I simply substitute in, and that is going to give me 6 times 42 (which is 252).
00:34:03.000 --> 00:34:12.400
So, the result from this composition of functions is 252, from evaluating h for -2; finding that result;
00:34:12.400 --> 00:34:20.000
evaluating g for that result (which was h(-2), which was 11); evaluating g for 11;
00:34:20.000 --> 00:34:26.600
finding that that was 42; and then evaluating f for that value.
00:34:26.600 --> 00:34:30.000
That concludes this session of Educator.com; thanks for visiting!