Consider the following function. f(x) = 1 ? x2/3 Find f(?1) and f(1). f(?1) = f(

Question

Consider the following function.

f(x) = 1 ? x2/3

Find f(?1) and f(1).

f(?1) =

f(1) =

Find all values c in (?1, 1) such that f?'(c) = 0. (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 Rolle's Theorem?

This contradicts Rolle's Theorem, since f is differentiable, f(?1) = f(1), and f?'(c) = 0 exists, but c is not in (?1, 1).

This does not contradict Rolle's Theorem, since f?'(0) = 0, and 0 is in the interval (?1, 1).

This contradicts Rolle's Theorem, since f(?1) = f(1), there should exist a number c in (?1, 1) such that f?'(c) = 0.

This does not contradict Rolle's Theorem, since f?'(0) does not exist, and so f is not differentiable on (?1, 1).

Nothing can be concluded.

Explanation / Answer

f(-1) = 1 ? (-1)^(2/3) = 0

and f(1) = 1 ? (1)^(2/3) = 0

Next we need to find the value for x=c , within the interval x E (-1,1) such that f '(c) = 0

First we find the derivative of f(x)

=> f '(x) = -(2/3)x^(-1/3)

when x = c

=> f '(c) = -(2/3)c^(-1/3)

next we need to solve f '(c) = 0

=> -(2/3)c^(-1/3) = 0

or c^(-1/3) = 0

=> c = 0 , This value of c is not possible as when c = 0 the function f(x) becomes not defined

=> There is no such value of c , within x E (-1 , 1) for which f '(c) = 0

Hence, c = Does Not Exist

As per Rolle's Theorem

If we have a function f(x) on x E [a,b] and,

(a) f(x) is continuous on x E [a,b]

(b) f(x) is differentiable on x E (a,b)

and (c) f(a) = f(b)

If the above three are fulfilled then there exist atleast one value x = c within the interval x E (a,b)
for which f '(c) = 0 holds true

Now we have the function
     
f(x) = 1 ? x^(2/3) , on x E (-1,1)

now f(x) is continuous on x E [a,b]

f(-1) = f(1) = 0

and f(x) is differentiable on all values of x within x E (-1,1) except for x = 0, so f(x) is not
differentiable at on of the x values within the given interval

=> f '(c) = 0 does not hold true.

So we conclude that :

This does not contradict Rolle's Theorem, since f '(0) does not exist, and so f is not
differentiable on (?1, 1).

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