Question

Consider the function $f\left(x\right)=\left(x^3-x\right)\left|x^2-6x+5\right|,\forall x\in R$ , then $f\left(x\right)$ is

• A. discontinuous at $x=1$
• B. discontinuous at $x=5$
• C. non differentiable at $x=1$
• D. non differentiable at $x=5$

Answer 1

Answer: (D)

$\begin{array}{l}f(x)=\left(x^3-x\right)|(x-1)(x-5)|\\ =x(x+1)(x-1)|(x-1)(x-5)|\\ f(x)=\{\begin{array}{c}\left(x^3-x\right)\left(x^2-6x+5\right)\\ -\left(x^3-x\right)\left(x^2-6x+5\right)\end{array}x\in (-\infty ,1)\cup (5,\infty )\\ \because \left(x^3-x\right),\left|x^2-6x+5\right|\text{ both are continuous }\forall x\in R\end{array}$
$\therefore f(x)$ is also continuous $\forall x\in R$
Now,
$f^{\prime }(x)=\{\begin{array}{cc}\left(x^3-x\right)(2x-6)+\left(x^2-6x+5\right)\left(3x^2-1\right)& x\in (-\infty ,1)\cup (5,\infty )\\ -\left(x^3-x\right)(2x-6)-\left(x^2-6x+5\right)\left(3x^2-1\right)& x\in (1,5)\end{array}$
Now, check differentiability
At $x=1$
$f^{\prime }\left(1^{-}\right)=\left(1^3-1\right)(2-6)+(1-6+5)(3-1)=0$
$f^{\prime }\left(1^{+}\right)=-(1-1)(2-6)-(1-6+5)(3-1)=0$
$\because f^{\prime }\left(1^{-}\right)=f\left(1^{+}\right)\therefore$ at $x=1$ differentiable
At $x=5$
$f^{\prime }\left(5^{-}\right)=(125-5)(10-6)+0$
$f^{\prime }\left(5^{+}\right)=-(125-5)(10-6)-0$
Clearly, $f^{\prime }\left(5^{-}\right)\ne f^{\prime }\left(5^{+}\right)$
$\therefore$ non-differentiable at $x=5$