Let f(x)=cases|4x^2-8x+5|, & if 8x^2-6x+10[4x^2-8x+5], & if 8x^2-6x+1

Question

Let $f(x)=\begin{cases}|4x^2-8x+5|, & \text{if } 8x^2-6x+1\ge0\[4x^2-8x+5], & \text{if } 8x^2-6x+1<0\end{cases}$, where $[\alpha]$ denotes the greatest integer $\le\alpha$. Then the number of points in $\mathbb{R}$ where $f$ is not differentiable is

Answer: 3

Accepted Answer

$8x^2-6x+1\ge0$ factors as $(4x-1)(2x-1)\ge0$, true for $x\in\left(-\infty,\frac14\right]\cup\left[\frac12,\infty\right)$, and $8x^2-6x+1<0$ on $\left(\frac14,\frac12\right)$.

So
$$f(x)=\begin{cases}4x^2-8x+5, & x\in\left(-\infty,\frac14\right]\cup\left[\frac12,\infty\right)\[4x^2-8x+5], & x\in\left(\frac14,\frac12\right)\end{cases}$$
On $\left(\frac14,\frac12\right)$, $4x^2-8x+5$ ranges continuously and $[4x^2-8x+5]=3$ for $x\in\left(\frac14,\frac{2-\sqrt2}{2}\right)$ and $=2$ for $x\in\left(\frac{2-\sqrt2}{2},\frac12\right)$.

So $f$ is non-differentiable at $x=\dfrac14,\ \dfrac{2-\sqrt2}{2},\ \dfrac12$ --- a total of 3 points.

Solutions · 1