Mean Value Theorem
Let f(x) = (x − 3)^{−2}.
Find all values of c in (2, 5) such that f(5) − f(2) = f '(c)(5 − 2).
This contradicts the Mean Value Theorem since f satisfies the hypotheses on the given interval but there does not exist any c on (2, 5) such that f '(c) = f(5) − f(2) |
5 − 2 |
.
This does not contradict the Mean Value Theorem since f is not continuous at x = 3. This does not contradict the Mean Value Theorem since f is continuous on (2, 5), and there exists a c on (2, 5) such that f '(c) = f(5) − f(2) |
5 − 2 |
.
This contradicts the Mean Value Theorem since there exists a c on (2, 5) such that f '(c) = f(5) − f(2) |
5 − 2 |
,
2 Answers By Expert Tutors
Shin C. answered • 07/04/20
UCLA Undergrad Expert/Enthusiast in Calculus, K-12, and SAT Math
Thanks for asking me another good question! Let's quickly recall what the Mean Value Theorem states:
As long f(x) is continous and differentiable on (a, b), then there must be a x-value c so a < c < b
and f ' (c) = [ f(b) - f(a) ] / ( b - a )!
Looking at the function, we see that f(x) = 1 / (x - 3) ^ 2. We can easily see that this function would be undefined at x = 3 because then the denominator would be 0, and we do not want that! But if we want to use the MVT from [2, 5], that would be impossible because x = 3 is undefined, and thus not continous nor differentiable at all points between all points of [2, 5]. Thus, because the conditions are not satisfied, we are unable to use the MVT!
Stephanie B. answered • 07/04/20
High School/College Algebra and Trig, Statistics, and Chemistry
Hi Hamoton,
Your post is a little jumbled, but let's look at two things and I think you will understand which option in your post is the correct answer.
First of all, look at the definition of the Mean Value Theorem:
If a function f is differentiable on an interval (a, b), and it is also continuous on [a, b], then there is at least one point c where the slope at c is equal to the secant line between a and b.
Writing that mathematically:
f'(c) = [f(b) - f(a)]/(b-a)
the slope at c = the slope of the secant line between a and b
In this problem, the function f(x) = (x-3)^{-2}, and hopefully you recognize that this function is not continuous in the interval [2, 5]. Specifically, it's discontinuous at x=3. <+++++ PART 1 of the answer
Secondly, if you calculate the slope of the secant line between 2 and 5, you'll find that it's -0.25.
If you try to find a point c on this function where the slope is equal to -0.25, you'll find that it is outside the interval [2, 5]... in other words, there's no point c in the interval [2, 5] where the Mean Value Theorem holds true. But, the Mean Value Theorem is not contradicted, because the Mean Value Theorem only applies to functions that are continuous on the interval <++++++ PART 2 of the answer
So hopefully now you can see that "This does not contradict the Mean Value Theorem since f is not continuous at x = 3"
This means, the result from Part 2 doesn't contradict the Mean Value Theorem, because the Mean Value Theorem only applies to continuous functions.
Hope that helps, and feel free to contact me if you have any more questions.
Steph
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Lois C.
For the Mean Value theorem to apply, the function must be continuous on the given interval, but at x = 3, the function has a discontinuity. The apparent choices of possible answers don't seem to cite this as a reason why the MVT doesn't work.07/04/20