INSTRUCTORS Raffi Hovasapian John Zhu

Start learning today, and be successful in your academic & professional career. Start Today!

• ## Related Books

 0 answersPost by Angela Zhang on December 7, 2014John, you are very cute! I like you <3 0 answersPost by Yu Han on July 24, 2014For example 1, how can you tell if it met these two conditions? 0 answersPost by Justin Powell on February 10, 2014Just for clarity:The Mean Value Theorem says that if a function f(x) is continuousand differentiable between two intervals x=a and x=b, then solving the function for these two values will give the coordinates {a,f(a)} and {b,f(b)}.Now if you draw a line between these two points, the slope will be:(f(b)-f(a))/(b-a) which is the rise over the run.The Mean Value Theoem states that at some point, {c,f(c)},on the graph of the function between the intervals x=a an x=b, the derivative or slope must be equal to (f(b)-f(a))/(b-a) at least once.So the derivative of f(x) at x=c is equal to:f'(c)=(f(b)-f(a))/(b-a)In the last example:the slope between the coordinates {-1,-1} and {5,125} equals 126/6 = 21=f'(c)=3c^2 c=sqrt(7).At f(sqrt7) the slope of the curve of the graph is parallel to the line between {-1,-1} and {5,125}.also for the MVT, f'(c)=0 is only true when f(a) and f(b) are equal (Rolles theorem).Sorry for overclarifying, but the video left out some key points. 1 answerLast reply by: Andrew MuMon Jan 6, 2014 8:30 AMPost by Willie Wang on January 13, 2013AT Example 4, why c cannot be negative sqr(7)ï¼Ÿ 1 answerLast reply by: Johnny ZamoraFri Jan 10, 2014 2:58 AMPost by James Xie on December 17, 2012What exactly does c stand for? Is it the Instant Rate Of Change (IROC? 0 answersPost by Ryan Menezes on November 18, 2012That's exactly what i want to know...... 1 answerLast reply by: Siyun LiuSat Jan 24, 2015 6:57 PMPost by Steve Denton on October 18, 2012At 5:00, what happened to the 1/2 in front of (c-1)^-1/2? rewritten then 1/ (2 sq rt (c-1))?????

### Mean Value Theorem

• There exists a point c, between aand b where the slope is defined by change in f(x)divided by change in x.
• Applying Mean Value Theorem
• Check conditions
• Take derivative
• Set derivative function =
• Replace xwith c
• Solve for c value

### Mean Value Theorem

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = ex on the interval [0, ln2]
• f′(x) = ex
• Continuous and differentiable everywhere
• f′(c) = [(f(ln2) − f(0))/(ln2 − 0)]
• f′(c) = [(eln2 − e0)/ln2]
• f′(c) = [(2 − 1)/ln2] = [1/ln2]
• Rolle's Theorem conditions are not met (f(a) ≠ f(b)), but the Mean Value conditions are met.
• f′(c) = ec = [1/ln2]
• lnec = ln[1/ln2]
c = ln[1/ln2]
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = sinx on the interval [0, 2π]
• f′(x) = cosx
• Continuous and differentiable everywhere
• f′(c) = [(cos2π− cos0)/(2π− 0)]
• f′(c) = [(1 − 1)/(2π)]
• f′(c) = 0
• cosc = 0
• There are many solutions, but only two in the interval
c = [(π)/2], [(3π)/2]
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = x3 − x2 + 1 on the interval [0, 1]
• f′(x) = 3x2 − 2x
• f′(c) = [(13 − 12 + 1 − 0)/(1 − 0)]
• f′(c) = 1
• 3c2 − 2c = 1
• 3c2 − 2c − 1 = 0
• c = [(2 ±√{4 − 4(3)(−1)})/(−6)]
• c = [(2 ±4)/(−6)]
• c = 1, [(−1)/3]
• Only one of these values is in the interval
c = 1
Find the mean value of f′(x) using the Mean Value Theorem for f(x) = 3x2 on the interval [1, 2]
• Polynomial. Continuous and differentiable.
• f′(x) = 6x
• f′(c) = [(3(22) − 3(12))/(2 − 1)]
f′(c) = 9
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [−1,1]
• f(x) is undefined at x = 0 and that is within the interval
• f(x) is not continuous on the interval
Mean Value Theorem cannot be applied
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [1, 2]
• f′(x) = −[1/(x2)]
• f′(c) = [([1/2] − [1/1])/(2 − 1)]
• f′(c) = −[1/2]
• −[1/2] = −[1/(c2)]
• [1/2] = [1/(c2)]
• c2 = 2
• c = ±√2
• Only one solution in the interval
c = √2
f(x) is continuous and differentiable on [a,b] and f(a) = f(b) and a ≠ b. Show that f′(x) has a root in the interval
• In order for some c to be a root of f′(x), f′(c) = 0
• Continuous and differentiable, so Mean Value Theorem can be used
• f′(c) = [(f(b) − f(a))/(b − a)]
• f′(c) = [(f(b) − f(b))/(b − a)]
• f′(c) = [0/(b − a)]
• f′(c) = 0
f′(c) = 0, so there is a root at x = c
Find at least one root of f′(x) on the interval [−1, 1] given f(x) = x2
• Polynomial. Continuous and differentiable everywhere
• f′(x) = 2x
• f′(c) = [(12 − (−12))/(1 − (−1))]
• f′(c) = 0
• 2c = 0
• c = 0
There's a root of f′(0)
f(x) is continuous and differentiable on [0,5]. Also, f(0) = 1 and f′(x) ≤ 3 in the interval. Find the largest possible value of f(5).
• f′(c) = [(f(5) − f(0))/(5 − 0)]
• f′(c) = [(f(5) − 1)/5]
• 5 f′(c) = f(5) − 1
• f(5) = 5 f′(c) + 1
• Plug in the largest value
• f(5) = 5(3) + 1
The largest possible value of f(5) is 16
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = √x on the interval [1, 4]
• f(x) is continuous so long as x ≥ 0
• f′(x) = [1/(2 √x)]
• Differentiable so long as x > 0
• Continuous and differentiable in the interval
• f′(c) = [(√4 − √1)/(4 − 1)]
• f′(c) = [1/3]
• [1/(2 √c)] = [1/3]
• 2√c = 3
• √c = [3/2]
c = [9/4]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Mean Value Theorem

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
• Mean Value Theorem 0:06
• Rolle's Theorem
• Mean Value Theorem Conditions
• Mean Value Theorem Definition
• Example 1 0:56
• Example 2 2:44
• Example 3 5:28
• Example 4 7:15