Let f(x) = (x − 3)^−2. Find all values of c in (2, 5) such that f(5) − f(2) = f '(c)(5 − 2).?

f(x) = (x - 3)⁻²
f '(x) = -2 (x - 3)⁻³
f '(c) = -2 (c - 3)⁻³

f(5) = (5 - 3)⁻² = 1/4
f(2) = (2 - 3)⁻² = 1

f(5) − f(2) = f '(c) (5 − 2)
(1/4) - 1 = -2 (c - 3)⁻³ (3)
-6 (c - 3)⁻³ = -3/4
(c - 3)⁻³ = 1/8
(c - 3)³ = 8
(c - 3)³ = 2³
c - 3 = 2
c = 5

As 5 is not in (2, 5).
Hence, no value of c is found.

(If c is in [2, 5], the answer is c = 5.)

f(x) = (x - 3)¯²

f(5) = (5 - 3)² = 4

f(2) = (2 - 3)² = 1

f’(x) = f(5) - f(2) / 5-2

f’(x) = -2(x - 3)¯³(1)

-2(x - 3)¯³ = 4 - 1/5 - 2

-2/(x - 3)³ = 3/3 = 1 → (x - 3)³ = -2

(c - 3)³ = -2

-(3 - c)³ = -2

3 - c = 8

-c = 8 - 3

c = -5

there is no ‘c’ in the (2, 5).

f(5) = 1/4. f(2) = 1.

We seek "c" such that
-3/4 = f'(c)(3) =>
f'(c) = -1/4.
f'(x) = -2/(x-3)^3. If this is -1/4, then
1/4 = 2/(x-3)^3 =>
(x-3)^3 = 8 =>
x = 5.

So there is no "c" in the open interval (2,5) that satisfies the conditions. And that's not surprising, because f(x) is NOT continuous in the interval (2,5).

f(5) = 1/4
f(2) = 1 .... m = [1/4 - 1]/[5 - 2]
m = -3/4 / 3 = -1/4
=============== =
f' = (-2)(x - 3)^-3
f'(c) = (-2)(c - 3)^-3 = -1/4
(c - 3)^-3 = 1/8
(c - 3)^3 = 8
c - 3 = 2
c = 5 <<< answer