Question

Let f(x) = (x-3)^(-2)
Find all values "c" in interval (1/4) such that
f(4) - f(1) = f'(c)*(4-1)

This is MVT.
ATTEMPT:
For f'(x), I got -2/(x-3)^3
For the secant slope from that line eqn above, I got 1/4.

Just by looking at f'(x), I reasoned that it will equal 1/4 when x = 1, or when c = 1. But c is not in the interval. Is the correct answer "DNE"? Let f(x) = (x-3)^(-2)
Find all values "c" in interval (1/4) such that
f(4) - f(1) = f'(c)*(4-1)

This is MVT.
ATTEMPT:
For f'(x), I got -2/(x-3)^3
For the secant slope from that line eqn above, I got 1/4.

Just by looking at f'(x), I reasoned that it will equal 1/4 when x = 1, or when c = 1. But c is not in the interval. Is the correct answer "DNE"? @Mathematics

Answer 1

The mean value theorem doesnt apply to this problem, because the function isnt continuous on the whole interval, and the function also isnt differentiable on the whole interval (for both look at x = 3). So the MVT doesnt hold in this case.

Answer 2

Oh I get it. Because x = 3 is stated to be in the interval (1,4), but DNE.

Answer 3

right.

Answer 4

WAIT!

Answer 5

Then what is c?

Answer 6

Apparently I put c = DNE, and I got this wrong.

Answer 7

its not guaranteed to exist, but it COULD exist

Answer 8

Sorry I wrote the interval wrong; (1,4)

Answer 9

hmm...im not too sure. let me think about it a little. maybe there is a such c in the interval [1,3) ?

Answer 10

its not guaranteed to exist, so you have to do some work to see if it does exist actually

Answer 11

Interval is this: (1, 4). THe only c I could think of is being c = x = 1, but this is not in the interval.

Answer 12

yeah, x = 1 is the only solution to f'(x)= 1/4. i dont know what else to say =/

Answer 13

i dont think it exists, no

Answer 14

My Apologies. I did not press Submit when I entered DNE. Thank you very much Joe and Sarah.