How do you find the absolute extrema of a function on a closed interval?

Question

Ex.

F(x)= 2(3-x), [-1, 2]

or

g(t)= (t^2)/(t^2 + 3) , [-1, 1]

Would a graphing calculator come in handy?

Answer 1

Let's take an example f(x) = 3x^2-2x+5 Interval: [-2,4]:

Step 1: differentiate
f '(x) = 6x-2
f ' '(x) = 6

Step 2: look for f ' (x)=0 (needed for relative extrema)
f ' (x) = 6x-2 = 0 ==> x=1/3

Step 3: verify f ' ' (1/3)<>0 (sufficient for relative extrema)
f ' ' (x) = 6>0 (correct) ==> relative low in x=1/3
f(1/3)=3*(1/3)^2-2*(1/3)+5= 14/3

Step 4: check the borders ( x=-2 and x=4)
f(-2)=3*(-2)^2-2*(-2)+5 = 21
f(4)=3*(4)^2-2*4+5 = 45

Step 5: compare the values and chose the highest / lowest
f(1/3) = 14/3 <== lowest
f(-2) = 21
f(4) = 45 <== highest

Step 6: label the points (if needed)
lowest in [-2,4]: P(1/3 / 14/3)
highest in [-2,4]: Q(4 / 45)

A graphic calculator could solve this too, but I guess for this example you'd be even faster without a calculator. Maybe you want to double-check your results with the calculator by drawing the function.

Answer 2

Absolute Extrema Calculator

Answer 3

Extrema Calculator

Answer 4

You have to find out the value of the function at the critical points
(f´(x)=0)m And the values at the extremes of the closed interval
The first has No critical points so the absolute extrema are at the end points
g(t) =1-3/(t^2+3)
g´(t) = -3/(t^2+3)^2(-2t)=6t/(t^2+3^2
t=0
g(0) =0
g(1)=g(-1) as g is an even function = 1/4
Max abs = 1/4 at the extremes and minimum abs at t=0

Answer 5

im pretty sure you need your calculator for that.