Question

Let f(x) = (x − 3)−2. Find all values of c in (1, 7) such that f(7) − f(1) = f '(c)(7 − 1). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) c = Based off of this information, what conclusions can be made about the Mean Value Theorem? This contradicts the Mean Value Theorem since f satisfies the hypotheses on the given interval but there does not exist any c on (1, 7) such that f '(c) = f(7) − f(1) 7 − 1 . 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 (1, 7), and there exists a c on (1, 7) such that f '(c) = f(7) − f(1) 7 − 1 . This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) 7 − 1 , but f is not continuous at x = 3. Nothing can be concluded.

Answer 1

Answer:

This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3

Step-by-step explanation:

The given function is

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

When we differentiate this function with respect to x, we get;

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

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)

This implies that;

$0.06-0.25=-\frac{2}{(c-3)^3} (6)$

$-0.19=-\frac{12}{(c-3)^3}$

$(c-3)^3=\frac{-12}{-0.19}$

$(c-3)^3=63.15789$

$c-3=\sqrt[3]{63.15789}$

$c=3+\sqrt[3]{63.15789}$

$c=6.98$

If this function satisfies the Mean Value Theorem, then f must be continuous on  [1,7] and differentiable on (1,7).

But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.