Question

Left hand derivative and right hand derivative of a function $f\left(x\right)$ at a point $x=a$ are defined as

$f^{\prime }\left(a^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{f\left(a\right)-f(a-h)}{h}=\underset{h\to 0^{-}}{lim}\frac{f\left(a\right)-f(a-h)}{h}=\underset{x\to a^{+}}{lim}\frac{f\left(a\right)-f\left(x\right)}{a-x}$ respectively

Let $f$ be a twice differentiable function. We also know that derivative of an even function is odd function and derivative of an odd function is even function.

The statement $\underset{h\to 0}{lim}\frac{f(-x)-f(-x-h)}{h}=\underset{h\to 0}{lim}\frac{f\left(x\right)-f(x-h)}{-h}$ implies that for all x $ϵR$

• A. $f$ is odd
• B. $f$ is even
• C. $f$ is neither odd nor even
• D. nothing can be concluded

Answer 1

The correct option is B $f$ is even

$\underset{h\to 0}{lim}\frac{f(-x)-f(-x-h)}{h}=\underset{h\to 0}{lim}\frac{f(-x-h)-f(-x)}{-h}$
$=f^{\prime }(-x)$ ......(i)
and $\underset{h\to 0}{lim}\frac{f\left(x\right)-f(x-h)}{-h}=-f^{\prime }\left(x\right)$ ......(ii)
from (i) and (ii) $f^{\prime }\left(x\right)$ is odd function and hence $f\left(x\right)$ is even function.