Q 19. - Calculus

Question

Every function of a single variable, $f\left(x\right)=x^n$, where n is a positive integer greater than $1$ will have a critical point at x = 0. For which values of n will there be a relative

minimum at x = 0? For which values of n will there be an

inflection point at x = 0? Are there any other possibilities

for the behavior of the function at x = 0?

Step-by-Step Solution

Answer

Critical point is calculated using derivative equals zero.

Step 1: Given Information

It is given that $f\left(x\right)=x^n$

Step 2: Explanation

Calculating derivative and equating with zero.

$n{x}^{n-1}=0$

It is possible only if $x=0$, critical point is $x=0$

$n$ is even then $f\left(x\right)=x^n>0$ for all $x\ne 0$

Zero is minimum value of $f\left(x\right)$

$x=0$ is minima when $n$ is even

When $n$ is odd $f\left(x\right)=x^n>0$,

if $x>0\text{ and }f\left(x\right)=x^n<0\text{ if }x<0$

Inflexion point is at $x=0$ for $n$ is odd function.