Let m and n be the number of points at which the function f(x)=maxx,x^3,x^5, x^21,x R, is …

Question

Let $m$ and $n$ be the number of points at which the function $f(x)=max{x,{x}^{3},{x}^{5},\ldots {x}^{21}},x\in R$, is not differentiable and not continuous, respectively. Then $m+n$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

Since $f$ is a maximum of finitely many continuous functions, $f$ is continuous everywhere, so $n=0$.

For $x\geq1$: $x^{21}\geq x^{19}\geq\cdots\geq x$. For $-1\le x\le0$: raising a number in $[-1,0]$ to a higher odd power brings it closer to $0$, so $x^{21}\ge x^{19}\ge\cdots\ge x$ there too. For $0<x<1$ and $x<-1$, the smallest power $x$ dominates the max. This gives:
$$f(x)=\begin{cases}x, & x<-1\ x^{21}, & -1\le x\le0\ x, & 0<x<1\ x^{21}, & x\ge1\end{cases}\qquad f'(x)=\begin{cases}1, & x<-1\ 21x^{20}, & -1\le x\le0\ 1, & 0<x<1\ 21x^{20}, & x\ge1\end{cases}$$

Checking the junctions: at $x=-1$, LHD $=1$, RHD $=21$ --- differ. At $x=0$, LHD $=0$, RHD $=1$ --- differ. At $x=1$, LHD $=1$, RHD $=21$ --- differ. So $f$ is non-differentiable at exactly $3$ points: $m=3$.

$m+n=3+0=3$

Solutions · 1