Holiday Blog.

Alrighttttt I already did one blog, so this is my second one and I guess i'm going to review some of chapter three. Which is when we started to learn max, mins, critical values and ect. So to start off extreme values are defined as maximums and minimums. Absolute Max is defined as the highest point on an interval
-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
-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:
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 there's a fraction only worry about the top of the fraction (only set the top=0).
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
ANSWER:
X=+/- 3

Example:
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. **Also plug in endpoints which are -1 and 2.
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 ABSOLUTE MIN
f(2)= 3(2)^4-4(2)^3= 16 ABSOLUTE MAX