Q. 8 - Calculus

Question

Sketch the graph of a function f that has an inflection point at $x=c$in such a way that the derivative $f\text{'}$has a local minimum at$x=c$. Add tangent lines to your sketch to illustrate that $f\text{'}$does have a local minimum at$x=c$. 

Step-by-Step Solution

Answer

 The function $f\left(x\right)=2x^3$has an inflection point at $x=0$and the first derivative $f\text{'}\left(x\right)=6x^2$has a local minimum at $x=0.$ 

Step 1. Given information.

 A function f that has an inflection point at $x=c$and the derivative has a local minimum at $x=c.$

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.

Consider a function $f\left(x\right)=2x^3.$

the first derivative is $f\text{'}\left(x\right)=6x^2.$

The second derivative is $f\text{'}\text{'}\left(x\right)=12x.$

the second derivative is zero for $x=0$so that

sign of f'' for the left of $x=0.$

$$\begin{gathered} f\text{'}\text{'}\left(1\right)=12 \\ f\text{'}\text{'}\left(1\right)>0 \end{gathered}$$

sign of f'' for the right of $x=0.$

$$\begin{gathered} f\text{'}\text{'}(-1)=-12 \\ f\text{'}\text{'}(-1)<0 \end{gathered}$$

so f is concave up for $x>0$and concave down for $x<0$ and the inflection  point at $x=0.$

Step 4. Graph of the function.

Plot the graph of the first derivative $f\text{'}\left(x\right)=6x^2.$

The function $f\left(x\right)=2x^3$has an inflection point at $x=0$and the first derivative $f\text{'}\left(x\right)=6x^2$has local minimum at $x=0.$