Question

If $f\left(x\right)$ is an even function and $f^{\prime }\left(x\right)$ exists, then $f^{\prime }\left(e\right)+f^{\prime }(-e)$ is

• A. $>0$
• B. $0$
• C. $\ge 0$
• D. $<0$

Answer 1

The correct option is B $0$

$f$ is called even function when $f(-x)=f\left(x\right)$

So $f(-x)-f\left(x\right)=0$

Differentiating at $x=e$ we get,

$f^{{}^{\prime }}(-e)+f^{{}^{\prime }}\left(e\right)=0$

$f^{{}^{\prime }}(-e)=-f^{{}^{\prime }}\left(e\right)$