Question

If we apply the Rolle's theorem to f(x)=x³-4x,x ∈[0,2], then c=.......,Select Proper option from the given options.
(a) √3
(b) 2
(c) 2/√3
(d) -2

Answer 1

Rolle's theorem is application of mean value theorem. in case of mean value theorem, f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a< c < b) such that $f'(c)=\frac{f(b)-f(a)}{b-a}$
if f'(c) = 0, it is said as Rolle's theorem,

so, f'(x) = 3x² - 4
now, f'(c) = 3c² - 4 = 0, where c belongs to [0, 2]
3c² - 4 = 0
c² = 4/3
taking square root both side,
c = ±2/√3 , but in [0, 2] only c = 2/√3

hence, option (c) is correct.

Answer 2

Hello,

Answer: c= 2/√3 (option c is correct)

Solution:

As Rolle's theorem is applicable on the given function,so we not to check whether Rolle's theorem is applicable for the given function.

We can directly calculate the values of c in given interval,i.e.

f'(c) =0

$f(x) = {x}^{3} - 4x \\ \\ f'(x) = 3 {x}^{2} - 4 \\ \\ f'(c) = 3 {c}^{2} - 4 = 0 \\ \\ 3 {c}^{2} = 4 \\ \\ {c}^{2} = \frac{4}{3} \\ \\ c = + - \frac{2}{ \sqrt{3} }$
c = 2/√3,which lies in the closed interval.

we have to discard c= -2/√3, as it is not in the closed interval

So,if Rolle's theorem is applicable,in the given interval,then there must a value,i.e. c=2/√3.

Hope it helps you.