Q. 30 - Calculus

Question

Find the critical points of each function f .Then use a graphing utility to determine whether f has a local minimum, a local maximum, or neither at each of these critical points.

$f\left(x\right)={\left(2x-1\right)}^5$

Step-by-Step Solution

Answer

The critical point is $x=\frac{1}{2}$ . The graph of the function is shown below .

Step 1. Given information .

Consider the given function $f\left(x\right)={\left(2x-1\right)}^5$ .

Step 2. Find the critical points .

To find the critical points differentiate the given function .

$$\begin{gathered} f\left(x\right)={\left(2x-1\right)}^5 \\ f\text{'}\left(x\right)=5{\left(2x-1\right)}^4 \\ =5{\left(2x-1\right)}^2{\left(2x-1\right)}^2 \end{gathered}$$

Further simplify .

Put $f\text{'}\left(x\right)=0$

$$\begin{gathered} {\left(2x-1\right)}^2 {\left(2x-1\right)}^2=0 \\ x =\frac{1}{2} , x=\frac{1}{2} \end{gathered}$$

Therefore the critical point is $x=\frac{1}{2}$ .

Step 3. Plot the graph .

The graph of the given function by using graphing utility is shown below.

From the given graph function f has local maximum because the turning point is on the positive axis .