Q.10 - Calculus

Question

In this problem, we will verify part (c) of Theorem 3.11. (a) For $f\left(x\right)=x^3$, show that $f\text{'}\left(0\right)=0$ and $f\text{'}\text{'}\left(0\right)=0$ while f does not have a local extremum at$x=0.$ 

(b) For $g\left(x\right)=x^4,$ show that $g\text{'}\left(0\right)=0 and g\text{'}\text{'}\left(0\right)=0$ while g has a local minimum at$x=0.$

(c) For $h\left(x\right)=-x^4,$show that $h\text{'}\left(0\right)=0 and h\text{'}\text{'}\left(0\right)=0$ while h has a local maximum at$x=0.$

(d) Explain how the parts (a)–(c) show that the second derivative test does not tell us anything when the second derivative at a critical point is zero. 

Step-by-Step Solution

Answer

(a) f is concave up for $x>0$ and concave down for $x<0$and the inflection  point at $x=0$but $f\text{'}\text{'}\left(0\right)=0$ so we can not predict whether the function has local minima or maxima or nither at $x=0.$

(b) the second derivative is positive for left and right of $x=0$ and the sign of the second derivative does not change at that point so the function does not have an inflection point and concave up for $x>0 \& x<0$and have local minima at $x=0.$

(c) Since the sign of the second derivative of function $g\left(x\right)$does not change at $x=0$so function does not have an inflection point  and have local maxima at $x=0.$

(d) In part (a)-(c), the change of sign of the second derivative cant be determined because the Value of the second derivative at $x=0$is zero so the second derivative test does not tell us anything when the second derivative is at a critical point is zero. 

Part (a) Step 1. Given information.

The given function is $f\left(x\right)=x^3.$$f\left(x\right)=x^3.$

the given derivatives are $f\text{'}\left(0\right)=0 \& f\text{'}\text{'}\left(0\right)=0.$

The given point is $x=0.$

Part (a) Step 2. Verification of local extremum.

The inflection point of the function is the point where the sign of the second derivative changes. 

Take function $f\left(x\right)=x^3.$

the first derivative of the function.

$f\text{'}\left(x\right)=3x^2.$

The second derivative of the function. 

$f\text{'}\text{'}\left(x\right)=6x.$

$f\text{'}\left(0\right)=0 \& f\text{'}\text{'}\left(0\right)=0$ for $x=0.$

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

$$\begin{gathered} f\text{'}\text{'}(-1)=-3 \\ f\text{'}\text{'}(-1)<0 \end{gathered}$$

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

$$\begin{gathered} f\text{'}\text{'}\left(1\right)=3 \\ f\text{'}\text{'}\left(1\right)>0 \end{gathered}$$

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

since $f\text{'}\text{'}\left(0\right)=0,$so we can not predict whether the function has local minima or maxima or nither at$x=0.$

plot the graph of $f\left(x\right)=x^3.$

Part (b) Step 1. Given information.

The given function is $g\left(x\right)=x^4.$

the given derivative is $g\text{'}\left(0\right)=0 \& g\text{'}\text{'}\left(0\right)=0.$

The given point is $x=0.$

Part (b) Step 2. Verification of local extremum.

The inflection point of the function is the point where the sign of the second derivative changes. 

Take function$g\left(x\right)=x^4.$

the first derivative of the function is$g\text{'}\left(x\right)=4x^3.$.

The second derivative of the function is $g\text{'}\text{'}\left(x\right)=12x^2.$

$g\text{'}\left(0\right)=0 \& g\text{'}\text{'}\left(0\right)=0$ for $x=0.$

sign of g' for the left of $x=0$

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

sign of g'' for the right of $x=0.$

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

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

the function has local minima at $x=0.$

Part (c) Step 1. Given information.

The given function is $h\left(x\right)=-x^4$

the given derivative is $h\text{'}\left(0\right)=0 \& h\text{'}\text{'}\left(0\right)=0$

The given point is $x=0.$

Part (c) Step 2. Verification of local extremum.

The inflection point of the function is the point where the sign of the second derivative changes. 

Take function $h\left(x\right)=-x^4.$

the first derivative of the function is$h\text{'}\left(x\right)=-4x^3$.

The second derivative of the function is$h\text{'}\text{'}\left(x\right)=-12x^2$. 

$h\text{'}\left(0\right)=0 \& h\text{'}\text{'}\left(0\right)=0$ for $x=0$

sign of h' for the left of $x=0$

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

sign of h'' for the right of $x=0$

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

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

the function has local maxima at $x=0.$

Part (d) Step 1. Explanation.

In part (a)-(c), the change of sign of the second derivative cant be determined because the Value of the second derivative at $x=0$is zero so the second derivative test does not tell us anything when the second derivative is at a critical point is zero.