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 [tex] f ' (x) < 0 [/tex], then function f(x) Decreasing in that interval. If [tex] f ' (x) > 0 [/tex], 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 [tex] f(x)=-x^3+9x-1 [/tex]
Step 1: Finding the derivative of the function:
[tex] f'(x)=-3x^2+9 [/tex]
Step 2: Set the[tex] f '(x) = 0 [/tex] and solve for x to get the critical numbers.
[tex] 0=-3x^2+9\\ \\-3x^2+9=0\\ -3x^2+9-9=0-9\\ -3x^2=-9\\ \\\frac{-3x^2}{-3} =\frac{-9}{-3}\\ \\ x^2=3\\ \sqrt{x^{2}} =\sqrt{3}\\ \\x=1.7(or)x=-1.7 [/tex]
Step 3: We need write the intervals based on the critical numbers
[tex] (-\infty, -1.7), (-1.7, 1.7),(1.7,\infty) [/tex]
Let us pick a Test point from the interval [tex] (-\infty, -1.7) [/tex] as -2
[tex] f ' (-2)=-3(-2)^2+9=-3(4)+9=-12+9=-3 [/tex]
The function will be decreasing in the interval[tex] (-\infty, -1.7) [/tex]
Let us pick a Test point from the interval [tex] (-1.7, 1.7) [/tex] as 0
[tex] f '(0)=-3(0)^2+9=0+9=9 [/tex]
The function will be increasing in the interval[tex] (-1.7, 1.7) [/tex]
Let us pick a Test point from the interval [tex] (1.7,\infty) [/tex] as 2
[tex] f'(2)=-3(2)^2+9=-3(4)+9=-12+9=-3 [/tex]
The function will be decreasing in the interval[tex] (1.7,\infty) [/tex].
Conclusion:
At [tex] x=-1.7 [/tex], function has a local minimum
[tex] f(-1.7)=-(-1.7)^3+9(-1.7)-1=-11.4 [/tex]
At [tex] x=1.7 [/tex], function has a local maximum
[tex] f(1.7)=-(1.7)^3+9(1.7)-1=9.4 [/tex]
