Question

Consider the function $f\left(x\right)=\left(x^{3} - x\right)\left|x^{2} - 6 x + 5\right|, \, \forall x\in R$ , 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)=\left\{\begin{array}{cc} \left(x^{3}-x\right)\left(x^{2}-6 x+5\right) \\ -\left(x^{3}-x\right)\left(x^{2}-6 x+5\right) \end{array} \quad x \in(-\infty, 1) \cup(5, \infty)\right. \\ \because\left(x^{3}-x\right),\left|x^{2}-6 x+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)=\left\{\begin{array}{cc} \left(x^{3}-x\right)(2 x-6)+\left(x^{2}-6 x+5\right)\left(3 x^{2}-1\right) & x \in(-\infty, 1) \cup(5, \infty) \\ -\left(x^{3}-x\right)(2 x-6)-\left(x^{2}-6 x+5\right)\left(3 x^{2}-1\right) & x \in(1,5) \end{array}\right. $
Now, check differentiability
At $x=1$
$f^{\prime}\left(1^{-}\right)=\left(1^{3}-1\right)(2-6)+(1-6+5)(3-1)=0$
$f^{\prime}\left(1^{+}\right)=-(1-1)(2-6)-(1-6+5)(3-1)=0$
$\because f^{\prime}\left(1^{-}\right)=f\left(1^{+}\right) \quad \therefore \quad$ 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, $f^{\prime}\left(5^{-}\right) \neq f^{\prime}\left(5^{+}\right)$
$\therefore$ non-differentiable at $x=5$