Question

Let $f\left(x\right)$ be a function satisfying the condition $f(-x)=f\left(x\right)$, for all real $x$. If $f^{\prime }\left(0\right)$ exists, then its value is

Answer 1

Since $f^{\prime }\left(0\right)$ exists,

$\therefore R{f}^{\prime }\left(0\right)=L{f}^{\prime }\left(0\right)$

$\Rightarrow \underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$

$=-\underset{h\to 0}{lim}\frac{f(-h)-f\left(0\right)}{h}=-\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$ $[\because f(-h)=f\left(h\right)]$

$\Rightarrow 2\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=0\Rightarrow 2f^{\prime }\left(0\right)=0$ $[\because f^{\prime }\left(0\right)$ exists$]$

$\Rightarrow f^{\prime }\left(0\right)=0$