Question

Number of points where $f\left(x\right)=(1-x)\left|x-x^2\right|+x$ is not differentiable is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

$f|x|=(1-x)|x(1-x)|+x$

$\left|x(1-x)\right|=\{\begin{array}{c}x(1-x)whenx(1-x)\ge 0\\ x(x-1)whenx(1-x)<0\end{array}\left|x(1-x)\right|=\{\begin{array}{c}x(1-x)whenx\in [0,1]\\ x(x-1)whenx\in (-\infty ,\infty )\cup (1,\infty )\end{array}$

$f\left(x\right)=\{\begin{array}{c}x{(1-x)}^2+xx\in [0,1]\\ =x(1+x^2-2x+1)\\ =x(x^2-2x+2)\\ x(x-1)(1-x)+xx\in (-\infty ,0)\cup (1,\infty )\\ =x(x^2+2x)=-x^3+2x^2\end{array}$

draw the graph of function $f\left(x\right)$.

$f^{\prime }\left(1^{-}\right)={\frac{d}{dx}(x^3-2x^2+2x)|}_{x=1}$

$={(3x^2-4x+2)|}_{x=1}$

$f^{\prime }\left(1^{+}\right)={\frac{d}{dx}(-x^3+2x^2)|}_{x=1}={(-3x^2+4x)|}_{x=1}=1$

$f^{\prime }\left(0^{+}\right)={(3x^2-4x+2)|}_{x=0}=2$

$f^{\prime }\left(0^{-}\right)={(-3x^2+4x)|}_{x=0}=0$

$f^{\prime }\left(0^{+}\right)\ne f^{\prime }\left(0^{-}\right)$ hence not differentiable at $x=0$

so $\left(B\right)1$ nor differentiability of point.