Taylor Rodriguez blog #8

This week we started chapter three. We began learning about extreme values, absolute max, absolute min, relative max, relative min, and critical numbers.
Extreme Values are defined as maximums and minimums
Absolute Max is defined as the highest point on an interval
** notice that absolute max is the highest point on a given interval not the highest point on the graph
Absolute min is defined as the lowest point on an interval
**notice that absolute min is the lowest point of a given interval not the lowest point on the graph
Relative Max (also referred to as local max) is defined as any max NOT on an interval
Relative Min (also referred to a local min) is defined as any min NOT on an interval
Critical Numbers are defined as any max or min typically referred to as x=c
When trying to solve problems concerning a critical number, extreme value, max, min, or horizontal tangents take the derivative and set equal to zero.

Example #1:
Find the critical values of f(x)= 9(x^2-3)/x^3
9x^2-27/x^3
Take the derivative
X^3(18x)-[(9x^2-27)(3x^2)]/ x^6
18x^4-[27x^4-81x^2]/x^6
18x^4-27x^4+81x^2/x^6
-9x^4+81x^2/ x^6
9x^2(-x^2+9)/X^6
9(-x^2+9)/x^4
Set equal to zero and solve
((when setting a fraction equal to zero you are really only setting what is in the numerator equal to zero))
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
X=+/- 3

Example #2:
Find the extrema of f(x)= 3x^4-4x^3 on the interval [-1,2]
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
X=0,1

To determine max or min on an interval plug into original and the biggest result will be the max
and the smallest result will be the min
f(-1)= 3(-1)^4-4(-1)^3=7
f(0)=3(0)^4-4(0)^3=0
f(1)= 3(1)^4 – 4(1)^3= -1 ((smallest so absolute min))
f(2)= 3(2)^4-4(2)^3= 16 ((largest so absolute max))