Question

Finding the absolute maximum and minimum of this function.

Answer 1

(x-2)(x-5)^3+4

Answer 2

On intervals [1,4] , [1,8] , and [4,9].

Answer 3

So the critical points I got were 11/4 and 5

Answer 4

x = 1 \[\left( 1-2 \right)(1-5)^3+4 = 68\]

Answer 5

x=11/4
\[(\frac{ 11 }{ 4 }-2)(\frac{ 11 }{ 4 }-5)^3+4=-4.54\]
x=4
\[(4-2)(4-5)^3+4=2\]

Answer 6

So for interval [1,4] the minimum should be 11/4 and no maximum, correct?

Answer 7

It's a closed interval, so there will be a maximum at x=1

Answer 8

x = 5
(5-2)(5-5)^3+4=4
x=8
(8-2)(8-5)^3+4=166
x=9
(9-2)(9-5)^3+4=452

Answer 9

Ok, so max=1 and min=11/4?

Answer 10

max = 68, min = -4.54
y-coords are max/min. x-coords are locations

Answer 11

Derp. That is why I missed them. Thanks for pointing that out @peachpi!

Answer 12

no problem

Answer 13

I got all of them right this time. Thanks again.