Question

For each function f that follows find all the x-values in the domain of f for which $f\text{'}\left(x\right)=0$and all the values for which $f\text{'}\left(x\right)$does not exist in later section we will call these values the critical points of f

localid="1648604345877" $$\begin{gathered} a)f(x)=x^3-2x \\ b)f\left(x\right)=\sqrt{x}-x \\ c)\hspace{0.17em}f(x)=\frac{1}{1+\sqrt{x}} \\ d)f\left(x\right)=\frac{x^2(x-1)}{{(x-2)}^2} \end{gathered}$$

Short answer

$$\begin{gathered} a)thederivativeiszerowhenx=\pm \sqrt{\frac{2}{3}} \\ b)thederivativeiszerowhenx=\sqrt[3]{4} \\ c)thederivativeiszerowhenx=0 \\ d)Thederivativeiszerowhenx=-0.443,7.007 \\ Andthefunctiondoesnotexistwhenx=2 \end{gathered}$$

Step-by-step solution

Given information

We are given functions

$$\begin{gathered} a)f(x)=x^3-2x \\ b)f\left(x\right)=\sqrt{x}-x \\ c)\hspace{0.17em}f(x)=\frac{1}{1+\sqrt{x}} \\ d)f\left(x\right)=\frac{x^2(x-1)}{{(x-2)}^2} \end{gathered}$$

Part a) Step 1: Find the derivative and the values of x f'(x)=0 and the values of x for which the derivative does not exist

We have,

$$\begin{gathered} \frac{d}{dx}(x^3-2x) \\ =3x^2-2 \end{gathered}$$

Now equate the derivative to zero

$$\begin{gathered} 3x^2-2=0 \\ {x}^2=\frac{2}{3} \\ x=\pm \sqrt{\frac{2}{3}} \end{gathered}$$

The function exist for all values of x

Part b) Step 1: Find the derivative and the values of x f'(x)=0 and the values of x for which the derivative does not exist.

We have,

$$\begin{gathered} \frac{d}{dx}(\sqrt{x}-x) \\ =\frac{1}{2}x\sqrt{x}-1 \end{gathered}$$

Now equate the derivative to zero

$$\begin{gathered} \frac{1}{2}x\sqrt{x}-1=0 \\ x\sqrt{x}=2 \\ {x}^3=4 \\ x=\sqrt[3]{4} \end{gathered}$$

Also the function exist for all values of x

Part c) Step 1:  Find the derivative and the values of x f'(x)=0 and the values of x for which the derivative does not exist.

We have,

$$\begin{gathered} \frac{d}{dx}\left(\frac{1}{1+\sqrt{x}}\right) \\ =-\frac{x\sqrt{x}}{{(1+\sqrt{x})}^2} \end{gathered}$$

Now we equate it to zero

$$\begin{gathered} -\frac{x\sqrt{x}}{{(1+\sqrt{x})}^2}=0 \\ thereforex=0 \end{gathered}$$

And the function will not exist when

$$\begin{gathered} 1+\sqrt{x}=0 \\ \sqrt{x}=-1 \end{gathered}$$

This is not possible hence the derivative exist for all values of x

Part d) Step 1: Find the derivative and the values of x , f'(x)=0  and the values of x for which the derivative does not exist.

We have,

$$\begin{gathered} \frac{d}{dx}\left(\frac{x^2(x-1)}{{(x-2)}^2}\right) \\ =\frac{x^4-8x^3+7x^2-2}{{(x-2)}^4} \end{gathered}$$

Now equate the derivative to zero

$$\begin{gathered} \frac{x^4-8x^3+7x^2-2}{{(x-2)}^4}=0 \\ {x}^4-8x^3+7x^2-2=0 \\ x=-0.433orx=7.007 \end{gathered}$$

And the derivative does not exist when

$$\begin{gathered} (x-2)=0 \\ x=2 \end{gathered}$$

Conclusion

$$\begin{gathered} a)thederivativeiszerowhenx=\pm \sqrt{\frac{2}{3}} \\ b)thederivativeiszerowhenx=\sqrt[3]{4} \\ c)thederivativeiszerowhenx=0 \\ d)Thederivativeiszerowhenx=-0.443,7.007 \\ Andthefunctiondoesnotexistwhenx=2 \end{gathered}$$