Lesson Relative Max or Relative Min of a function using Derivatives

Source code of 'Relative Max or Relative Min of a function using Derivatives'

This Lesson (Relative Max or Relative Min of a function using Derivatives) was created by by Nate(3500)

First, find the derivative of the function. That will describe the slope of the tangent line. When the tangent line has a slope of zero, the line is horizontal (basically located at "turning" points on the graphed line). A horizontal line is located at the relative max or relative min. Sometimes, we can not do this.
Example:
f(x) = x^3 + 2x - 1
f`(x) = 3x^2 + 2
0 = 3x^2 + 2
-2/3 = x^2
Imaginery....
Example:
f(x) = 2x^3 + 4x^2 + 3
f`(x) = 6x^2 + 8x
0 = 6x^2 + 8x
0 = x^2 + 8x/6
64/144 = (x + 8/12)^2
-8/12 +- 8/12 = x
x = 0 and x = -4/3
f(0) = 3
f(-4/3) = 2(-4/3)^3 + 4(-4/3)^2 + 3 = -128/27 + 192/27 + 81/27 = 145/27
Relative Min: 3 at x = 0
Relative Max: 17/3 or 5.370370 .... at x = -4/3
{{{graph(300,300,-2,2,-10,10,3,145/27,2x^3 + 4x^2 + 3)}}}
Example:
f(x) = x^2 - 2x + 1
f`(x) = 2x - 2
0 = 2x - 2
x = 1
f(1) = 1 - 2 + 1 = 0
Relative Min: 0 at x = 1
{{{graph(300,300,-5,5,-5,5,0,x^2 - 2x + 1)}}}
Example:
f(x) = x^5 - (5/3)x^3 - 1
f`(x) = 5x^4 - 5x^2
0 = 5x^2(x^2 - 1)
x = 0
x = -1
x = 1
f(0) = -1
f(-1) = -1/3
f(1) = -5/3
Relative Max: -1/3 at x = -1
Relative Min: -5/3 at x = 1
{{{graph(300,300,-2,2,-5,5,-1/3,-5/3,x^5 - (5/3)x^3 - 1)}}}
second Derivatives: ~> refer to last example
f'(x) = 5x^4 - 5x^2
f''(x) = 20x^3 - 10x
Just find the derivative of the function again.
"Turning" points at x: -1 and 1
f''(-1) = -20 + 10 = -10
-10 < 0 Maximum
f''(1) = 20 - 10 = 10
10 > 0 Minimum
Here is how it is done. After you find the second Derivatent of the numbers:
if greater than zero .... f''(x) > 0 .... it is a minimum
if equal to zero .... f''(x) = 0 .... it is either a maximum, minimum, or neither
if less than zero .... f''(x) < 0 .... it is a maximum