Question

Value of $c$ of Lagranges mean theorem for
$f\left(x\right)=2+x^3$ if $x\le 1$
$=3x$ if $x>1$ on $[-1,2]$ is

• A. $\pm \frac{\sqrt{5}}{3}$
• B. $\pm \frac{\sqrt{3}}{2}$
• C. $\pm \frac{\sqrt{2}}{5}$
• D. $\pm \frac{3}{\sqrt{5}}$

Answer 1

The correct option is A $\pm \frac{\sqrt{5}}{3}$

Lagrange's mean value theorem states that if $f\left(x\right)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ then there exists some $c$ between $a$ and $b$ such that $f^{{}^{\prime }}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Given $f\left(x\right)=2+x^3$ if $x\le 1$

$=3x$ if $x>1$ and $[a,b]=[-1,2]$

$⟹f^{{}^{\prime }}\left(x\right)=3x^2$if $x\le 1$

$=3$ if $x>1$

Therefore, $f^{{}^{\prime }}\left(c\right)=\frac{\left(3\right(2\left)\right)-(2+(-1{)}^3)}{2-(-1)}$

$⟹3c^2=\frac{5}{3}$

$⟹c^2=\frac{5}{9}$

$⟹c=\pm \frac{\sqrt{5}}{3}$