Mathematics, 16.07.2019 04:20 emmamerida

Let f(x) = (x − 3)−2. find all values of c in (2, 5) such that f(5) − f(2) = f '(c)(5 − 2). (enter your answers as a comma-separated list. if an answer does not exist, enter dne.)

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Answer from: accacia19

The answer does not exist.

Note - The statement has typing mistakes. Correct form is presented below:

Let $f(x) = (x-3)^{-2}$ . Find all values of in (2, 5) such that $f(5) - f(2) = f'(c) \cdot (5-2)$ . (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

In this question we should use the Mean Value Theorem, which states that given a secant line between points A and B, there is at least a point C that belongs to the curve whose derivative exists.

We begin by calculating f(2) and f(5) :

$f(2) = (2-3)^{-2}$

f(2) = 1

$f(5) = (5-3)^{-2}$

f(5) = 1

And the slope of the derivative is:

$f'(c) = \frac{f(5) - f(2)}{5-2}$

f'(c) = 0

Now we find the derivative of the function:

$f'(x) = -2\cdot (x-3)^{-3}$

$-2\cdot (x-3)^{-3} = 0$

-2 = 0 (ABSURD)

Hence, we conclude that the answer does not exist.

We kindly invite to see this question on Mean Value Theorem: link

Answer from: nickonicyetlan

Step-by-step explanation:

This is the Mean Value Theorem for Calculus. The formula for that, at least the one I use for it, resembles yours, but I don't simplify it down quite as far, as I find it easier to work with:

$f'(c)=\frac{f(b)-f(a)}{b-a}$

That says that the derivative of the function at some value of c within a < c < b

is equal to the slope of the function on the interval.

We need f'(c), f(b) and f(a) to solve this. First the derivative. I have to say that I am assuming the function is

$f(x)=(x-3)^{-2}$

If so, the derivaitve of this function is

$f'(x)=\frac{-2}{x-3}$

Next, f(b). We plug in a 5 for x in the function to get that f(b) = .25

Next, f(a). We plug in a 2 for x in the function to get that f(a) = 1

Filling in the MVT:

$\frac{-2}{x-3}=\frac{.25-1}{5-2}$

Cross multiply to get

-.75(x - 3) = -6 and

-.75x + 2.25 = -6 so

-.75x = -8.25 and

x = 11 (that is the same as c = 11). Now we can find the y value that corresponds to that x value by subbing 11 into x in the function:

$f(11)=(11-3)^{-2}$

$f(11)=\frac{1}{64}$

Therefore, the value of c where the slope of the function is the same as the slope of the derivative is at (11, 1/64)

Let f(x) = (x − 3)−2. find all values of c in (2, 5) such that f(5) − f(2) = f '(c)(5 − 2). (enter y

Answer from: Quest

answer: ur v dumb

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

the answer would be 7

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Let f(x) = (x − 3)−2. find all values of c in (2, 5) such that f(5) − f(2) = f '(c)(5 − 2). (enter y...

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