Question

The value of ${}^{\prime }{c}^{\prime }$ by Rolle's Theorem for which $f\left(x\right)=1-\sqrt[3]{3x^4}$ in $[-1,1]$ is

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is A $0$

Value of $c$ by Rolle’s theorem. $f\left(x\right)=1-\sqrt[3]{3x^4}$ in $[-1,1]$

$⟶$ Rolle’s theorem states that if a fraction $f\left(x\right)$ is antinuorson the interal $[a,b]$ and differentiable on the interval $(a,b)$ and if $=f\left(a\right)=f\left(b\right)$ then the these exists $Cϵ(a,b)$ such that $f\left(c\right)=0$

$⟶$ Here, $f\left(x\right)=1-\sqrt[3]{3x^4}$

$f(-x)=1-\sqrt[3]{3{(-1)}^4}=1-1=0$

$f\left(1\right)=1-\sqrt[3]{31^4}=1-1=0$

$f^{‘}\left(1\right)=\frac{d}{dx}{x}^{4/3}=-\frac{4}{3}{x}^{1/3}$

$⟶$ therefore,

$f^{‘}\left(c\right)=0$

$\therefore -\frac{4}{3}{c}^{7/3}=0$

$\therefore c^{7/3}=0$

$\therefore c=0$

$⟶$ so answer is $0$.