Question

Find the absolute maximum and absolute minimum values of the function

f(x)=(x−2)(x−7)3+7

on each of the indicated intervals.

(a) Interval = [1,4].

1. Absolute maximum =
2. Absolute minimum =

(b) Interval = [1,8].

1. Absolute maximum =
2. Absolute minimum =

(c) Interval = [4,9].

1. Absolute maximum =
2. Absolute minimum =

Answer 1

Hey.

Answer 2

You can find the derivative of this right?

Answer 3

yeah

Answer 4

Is that \[f(x)=(x-2)(x-7)^3+7\]

Answer 5

yeah

Answer 6

the derivative is (x-7)^2 (4 x-13)

Answer 7

we are finding the derivative to find the critical numbers
you will need to solve f'(x)=0 for x
we don't have to find when f' dne since this is a polynomial and polynomials are differentiable everywhere

Answer 8

i'm going to check your derivative before we move on

Answer 9

looks fabulous

Answer 10

ok now that is very first step in problem like
i will get you the recipe

1) find derivative
2) check if f is continuous on the domain given (don't worry here; polys are continuous everywhere)
3) find for what x is the derivative zero solve f'=0 for x
4) find for what x is the derivative not existing (don't worry here; polys are differentiable everywhere)
5) then evaluate f(endpoints) and f(critical numbers)
6) determine which values are highest and which are lowest in 5 for answer

Answer 11

so we need to find when (x-7)^2(4x-13)=0

Answer 12

when is x-7=0?
when is 4x-13=0?

Answer 13

oh also on number 5 you only want to plug in critical numbers that are between the endpoints

Answer 14

so whats the first thing after the derivative?

Answer 15

we are trying to find when the derivative is zero

Answer 16

that is what i set the derivative equal to zero and asked you to solve it

Answer 17

do you know how to solve (x-7)^2(4x-13)=0?

Answer 18

recall if you have a*b=0
this implies a=0 or b=0
or both =0

Answer 19

so x-7=0 or 4x-13=0
so x=? or x=?

Answer 20

i just did that and theres no number that would make 4x-13 equal to 0, and the answer for the other one is 7

Answer 21

you can solve 4x-13=0

Answer 22

come on you are taking calculus you have to recall some algebra for it

Answer 23

if you want to get the x by itself
you need to do two operations to both sides

Answer 24

and i say that because there is subtraction to undo
and also multiplication

Answer 25

ok, the number is 3.25

Answer 26

4x-13=0
+13 +13 <--add 13 to both sides
4x =13
-- -- <---divide 4 on both sides
4 4

x=13/4


you also found the other critical number x=7

we have two critical numbers in all


we will found out which ones matter when we look at the intervals

Answer 27

So the first function is defined [1,4]

we only care about the function from x=1 to x=4

Answer 28

but guess what x=7 doesn't occur between there so we don't care about x=7 for the first part.

Answer 29

ohh but the second one does right?

Answer 30

now go to number 5 of my recipe
evaluate
f(endpoints) and f(critical numbers)
make sure to only plug in critical number we care about for this one

Answer 31

yes and we will get to that one

Answer 32

you can go ahead and evaluate f(7) and f(13/4)
but we will not use f(7) for the first one

Answer 33

so what is f(1) and f(4) and f(13/4)

Answer 34

ok give me a sec

Answer 35

(x-7)^2 (4 x-13)

when we use 1, we get -324
when we use 4, we get 27
when we use 13/4, we get 0

Answer 36

did you find f(1) and f(4) and f(13/4)?

Answer 37

i need those values

Answer 38

yeah

Answer 39

i don't care where the derivative it at that number

Answer 40

i care about where they occur on the graph
i'm trying to determine the max and the min

Answer 41

ok can you give me those values...

Answer 42

Recall f(x)=(x-2)(x-7)^3+7
to find f(1) you replace the x's with 1 like so
f(1)=(1-2)(1-7)^3+7

Answer 43

then evaluate and simplify

Answer 44

ohh my bad i took them in the derivative

Answer 45

(x−2)(x−7)^3+7
at 1, it is 223
at 4, its -47
at 13/4, -58.91

Answer 46

groovy
now which of those is the highest value (absolute max)
which of those is the smallest value (absolute min)

Answer 47

223 absolute max and -58.91 absolute min

Answer 48

and i like exact numbers so instead of -58.92 i would say -15083/256

Answer 49

ok anyways you should be able to do the other problems
they are the same exact way

Answer 50

thank you

Answer 51

go ahead and do the 2nd one and let me know what you get

Answer 52

also the 2nd one is tons easier because we already did most of the work

Answer 53

you don't need to find f' again
you don't need to find the critical numbers x=13/4 or x=7

Answer 54

you also don't need to evaluate f(1) again

Answer 55

or f(13/4)

Answer 56

ok will do

Answer 57

now i said you don't need to find them but you still need to use the info

Answer 58

i found the other two
2) max: 223 and min: -58.91
3) max: 63 and min: -47

Answer 59

looks good

Answer 60

great job proto