this week was mostly going over optimization in preparation for the quiz on wednesday..friday we got our study guides for the exam.

Some of the stuff on the study guides are tangent lines and using the first/second derivative test.

A Tangent Line is a line which locally touchesa curve at one and only one point.• The slope-intercept formula for a line is y = mx + b,where m is the slope of the line and b is the y-intercept.

• The point-slope formula for a line is y – y1 = m (x – x1).This formula uses a point on the line, denoted by (x1, y1),and the slope of the line, denoted by m, tocalculate the slope-intercept formula for the line.

• The first derivative is an equation for the slope of a tangentline to a curve at an indicated point.The equation for the slope of the tangent line tof(x) = x2 is f '(x), the derivative of f(x).f(x) = x2f '(x) = 2x (1)Therefore, at x = 2, the slope of the tangent line is f '(2).f '(2) = 2(2)= 4

Now , you know the slope of the tangent line, which is 4.All that you need now is a point on the tangent line to beable to formulate the equation. To find that point, simply plugthe coordinate of the shared point into the original equation, this gives you (2,4)The only step left is to use the point (2, 4) and slope, 4,in the point-slope formula for a line. Therefore: Y-4=4(x-2)

TO FIND MAX AND MINS

1. first derivative test

2. plug the critical values into origional function to get y-values.

3. plug endpoints in to origional function to get y-values.

4. highest y-value is absolute max.

5. lowest y-value is absolute min.
-absolute maxs or mins or written as a point or simply as the y-value.

For example:find the absolute max or min of f(x)=3x^4-4x^ on [-1,2].f1(x)=12x^3-12x^2=012x^2(x-1)=0x=1,0(-1, 0)U(0,1)U(1,2)f1(-.5)=-ve f1(.5)=-ve f1(1.5)=+ve
min @x=13(1)^4-4(1)^3=-1 (1,-1)3(-1)^4-4(-1)^3=-7 (-1,-7)3(2)^4-4(2)^3 (2,16)-32=16
abs min:(1,-1)or -1 -1 at x=1


i still don't fully understand optimization as a whole, anyone want to explain from the beginning?