Question

The value of c for the mean-value theorem for f(x)=x³ in [-1,1] is .......,Select Proper option from the given options.
(a) ± 1/√3
(b) ± √3
(c) ± 1
(d) 0

Answer 1

please see the attachment

Answer 2

Answer:

$c = \ \±\frac{1}{\sqrt{3}}$

Step-by-step explanation:

In this question

We have been given that

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

So, f(x) is continuous in [-1, 1] and differential in [-1, 1]

Value of [a, b] = [-1, 1]

Then f(a) = $(-1)^{3}$ = -1

         f(b) = $(1)^{3}$ = 1

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

                  f(c) = $\frac{1 - (-1)}{1-(-1)}$ = 1

Now, f'(x) = 3$x^{2}$

         f'(c) = 3$c^{2}$ = 1

                   $c^{2}$ = $\frac{1}{3}$

                   $c = \ \±\frac{1}{\sqrt{3}}$

Hence, the value of $c = \ \±\frac{1}{\sqrt{3}}$