Q. 9 - Calculus

Question

Show that for $f\left(x\right)=x^6$we have $f\text{'}\text{'}\left(0\right)=0$but the point $x=0$is not an inflection point of f. 

Step-by-Step Solution

Answer

The f'' is positive for left and right of $x=0$ and the sign of second derivative does not change at $x=0,$ so function $f\left(x\right)=x^6$does not have inflection point at $x=0$  and function is concave up for $x>0 \& x<0.$

Step 1. Given information.

 The given function is $f\left(x\right)=x^6.$

The given second derivative is $f\text{'}\text{'}\left(0\right)=0$

given point is $x=0.$

Step 2. inflection point of the function.

The inflection point of a function is the point where the concavity of the function is changes and the second derivative of the function state the concavity of the function so the inflection point of the function can determine the point where the sign of second derivative changes.

Suppose function f has critical point at $x=c$so that $f\text{'}\left(c\right)=0$then,

if $f\text{'}\text{'}\left(c\right)$is positive then f will have local minima at $x=c.$

if $f\text{'}\text{'}\left(c\right)$is negative then f will have local maxima at $x=c.$

if $f\text{'}\text{'}\left(c\right) =0$then we cannot determine whether the function has local maxima or minima at $x=c.$

Step 3. concavity and inflection point of the function.

If the second derivative f'' of a function f is Positive then f'  is increasing on the interval, and if f' is increasing then f will concave up on interval.

Therefore f is concave up or down can be checked by the sign of its second derivative.

Now take function $f\left(x\right)=x^6.$

the first derivative is $f\text{'}\left(x\right)=6x^5.$

The second derivative is $f\text{'}\text{'}\left(x\right)=30x^4.$

the $f\text{'}\text{'}\left(0\right)=0$ for $x=0$so that

sign of f'' for the left of $x=0.$

$f\text{'}\text{'}\left(1\right)=30$

sign of f'' for the right of $x=0.$

$f\text{'}\text{'}(-1)=30$

the second derivative is positive for left and right of $x=0$ and the sign of the second derivative does not change at $x=0,$ so function does not have  inflection  point at $x=0$ and concave up for $x>0 \& x<0.$