Question

Find the local and global extrema to the nearest tenth for the graph of ƒ(x) = –x3 + 9x – 1

Answer 1

General Idea:

In order to find the local extrema of a function f(x), we need to do the below steps:

(i) Find the first derivative of the given function

(ii) Set the first derivative to zero and solve for x to identify the critical numbers.

(iii) Draw a number line plotting the critical numbers in it, then pick a test point from each of the intervals to check whether the function is increasing or decreasing.

(iv) If
`f ' (x) < 0`, then function f(x) Decreasing in that interval. If
`f ' (x) > 0`, then the function f(x) Increasing in that interval. Based on this information we can identify the local extrema's.

Applying the concept:

Given function
`f(x)=-x^3+9x-1`

Step 1: Finding the derivative of the function:

`f'(x)=-3x^2+9`

Step 2: Set the
`f '(x) = 0` and solve for x to get the critical numbers.

`0=-3x^2+9\\ \\-3x^2+9=0\\ -3x^2+9-9=0-9\\ -3x^2=-9\\ \\(-3x^2)/(-3) =(-9)/(-3)\\ \\ x^2=3\\ \sqrt{x^(2)} =√(3)\\ \\x=1.7(or)x=-1.7`

Step 3: We need write the intervals based on the critical numbers

`(-\infty, -1.7), (-1.7, 1.7),(1.7,\infty)`

Let us pick a Test point from the interval
`(-\infty, -1.7)` as -2

`f ' (-2)=-3(-2)^2+9=-3(4)+9=-12+9=-3`

The function will be decreasing in the interval
`(-\infty, -1.7)`

Let us pick a Test point from the interval
`(-1.7, 1.7)` as 0

`f '(0)=-3(0)^2+9=0+9=9`

The function will be increasing in the interval
`(-1.7, 1.7)`

Let us pick a Test point from the interval
`(1.7,\infty)` as 2

`f'(2)=-3(2)^2+9=-3(4)+9=-12+9=-3`

The function will be decreasing in the interval
`(1.7,\infty)`.

Conclusion:

At
`x=-1.7`, function has a local minimum

`f(-1.7)=-(-1.7)^3+9(-1.7)-1=-11.4`

At
`x=1.7`, function has a local maximum

`f(1.7)=-(1.7)^3+9(1.7)-1=9.4`