Q 4 - Calculus

Question

Finding critical points: For each of the following functions $f$, find all of the x-values for which $f\text{'}\left(x\right)=0$ and all of the x-values for which $f\text{'}\left(x\right)$ does not exist.

$f\left(x\right)={\left(1-x^4\right)}^7$

Step-by-Step Solution

Answer

The points for which $f\text{'}\left(x\right)=0$ are:-

$0,1,-1$.

There is no points for which $f\text{'}\left(x\right)$ does not exist.

Step 1. Given Information

We have given the following function:-

$f\left(x\right)={\left(1-x^4\right)}^7$

Step 2. Find the derivative of the given function

Firstly we will find the derivative, then we will find the required points.

We have given the following function:-

$f\left(x\right)={\left(1-x^4\right)}^7$.

Use the power rule and chain rule to differentiate this function, then we have:-

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}{\left(1-x^4\right)}^7 \\ \Rightarrow f\text{'}\left(x\right)=7{\left(1-x^4\right)}^6\frac{d}{dx}\left(1-x^4\right) \\ \Rightarrow f\text{'}\left(x\right)=7{\left(1-x^4\right)}^6\left(-4x^3\right) \end{gathered}$$

Step 3. Put f ' x = 0 .

We find that:-

$f\text{'}\left(x\right)=7{\left(1-x^4\right)}^6\left(-4x^3\right)$.

Now put $f\text{'}\left(x\right)=0$, then we have:-

$$\begin{gathered} 7{\left(1-x^4\right)}^6\left(-x^4\right)=0 \\ \Rightarrow {\left(1-x^4\right)}^6\left(-4x^3\right)=0 \\ \Rightarrow {\left(1-x^4\right)}^6=0 and \left(-4x^3\right)=0 \\ \Rightarrow 1-x^4=0 and x^3=0 \\ \Rightarrow x^4=1 and x=0 \\ \Rightarrow x=\pm 1 and x=0 \end{gathered}$$

Step 4. The values for which f ' ( x ) does not exist.

We find that :

$f\text{'}\left(x\right)=7{\left(1-x^4\right)}^6\left(-4x^3\right)$.

We know that a function does not exist where the denominator is equal to zero. But here denominator is constant which is 1. 

So it cannot be equal to zero.

So we can conclude that there is no point for which $f\text{'}\left(x\right)$ does not exist.