If f(x)=cases1|x|, & |x|1^2+b, & |x|

Question

If $f(x)=\begin{cases}\dfrac{1}{|x|}, & |x|\ge1\ax^2+b, & |x|<1\end{cases}$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively

Accepted Answer

$$f(x)=\begin{cases}-\frac1x, & x\le-1\ax^2+b, & -1<x<1\\frac1x, & x\ge1\end{cases}\qquad f'(x)=\begin{cases}\frac1{x^2}, & x<-1\2ax, & -1<x<1\-\frac1{x^2}, & x>1\end{cases}$$

Differentiability at $x=1$: $2a=-1\Rightarrow a=-\dfrac12$.

Continuity at $x=1$: $a+b=1\Rightarrow b=1-\left(-\dfrac12\right)=\dfrac32$.

So $a=-\dfrac12,\ b=\dfrac32$.

Solutions · 1