Question

Let $f\left(x\right)=|x-a|\varphi \left(x\right)$ , where $\varphi$ is a continuous function and $\varphi \left(a\right)\ne 0$. Then

• A. $f^{\prime }\left(a^{+}\right)=\varphi \left(a\right)$
• B. $f$ is differentiable at $x=a$
• C. $f^{\prime }\left(a^{+}\right)=-\varphi \left(a\right)$
• D. $f^{\prime }\left(a^{-}\right)=-\varphi \left(a\right)$

Answer 1

The correct option is D $f^{\prime }\left(a^{-}\right)=-\varphi \left(a\right)$

$f\left(x\right)=(x-a)\varphi \left(x\right)x>a$
$-(x-a)\varphi \left(x\right)x<a$
$f^{{}^{\prime }}\left(x\right)=\varphi \left(x\right)+(x-a){\varphi }^{{}^{\prime }}\left(x\right)x>a$
$-\varphi \left(x\right)-(x-a){\varphi }^{{}^{\prime }}\left(x\right)x<a$
$f^{{}^{\prime }}\left(a^{+}\right)=\varphi \left(a\right)$
$f^{{}^{\prime }}\left(a^{-}\right)=-\varphi \left(a\right)$
$f\left(x\right)$ is not differentiable at $x=a$