Question

@jim_thompson5910

Determine whether the Mean Value Theorem applies to f(x)=3x-x^2 on the interval [2,3]. If the Mean Value Theorem applies, find all value(s) of c in the interval such that f'(c) - f(3)-f(2)/3-2. If it does not apply, state why.

Answer 1

ok, the first condition we need to check is to see if f(x) is continuous on [2,3]

Answer 2

because if it is not, then the MVT won't work

Answer 3

It is

Answer 4

yes because it's continuous everywhere (it's a polynomial)

Answer 5

Yup

Answer 6

also, we can easily determine that f(x) is differentiable everywhere

so f(x) is differentiable on (2,3)

Answer 7

so the MVT applies in this case

Answer 8

now you need to compute (f(3) - f(2))/(3 - 2)

Answer 9

what do you get?

Answer 10

I gotttt -2

Answer 11

me too

Answer 12

I set f'(x) equal to that and solve for x.. right?

Answer 13

now you need to calculate f ' (x)

then set it equal to -2 and solve for x to get the value of c such that f ' (c) = -2

Answer 14

yes correct

Answer 15

x=5/2

Answer 16

correct

Answer 17

so c = 5/2

Answer 18

Now what

Answer 19

you found the value of c such that

\[\Large f ' (c) = \frac{f(3)-f(2)}{3-2}\]

so you're done with the problem

Answer 20

c = 5/2

Answer 21

Oh yaya lol

Answer 22

and again, the MVT applies because f(x) is continuous on [2,3] AND differentiable on (2,3)

Answer 23

so that's why it works

Answer 24

I didnt get full credit on this one:

Find the minimum and maximum values of f(x)=x^2-2x+1 on the interval [0,3]

I got x=1 and yeah I could only find a min at x=1?

Answer 25

x = 1 is where the min happens, but you need to plug it back into f(x) to get the actual min value

Answer 26

also, you need to test the endpoints

Answer 27

I did that.

He wants a aximum too.

Answer 28

The min is that (1,0)

Answer 29

with closed intervals, the abs min/max could be at the endpoints

Answer 30

yes the min is at (1,0)

the min value is 0

Answer 31

Min value is 1

Answer 32

no the min value is 0 because that is the result of f(1)

Answer 33

now you need to calculate f(0) and f(3)

Answer 34

I dont understand lol

Answer 35

the endpoints of the interval are at x = 0 and x = 3

this is because the interval is [0,3]

Answer 36

f(x)=x^2-2x+1

f(0)=0^2-2(0)+1

f(0) = 1

Answer 37

f(x)=x^2-2x+1

f(3)=(3)^2-2(3)+1

f(3) = 4

Answer 38

notice how x = 3 produces the largest output

f(3) = 4 is the largest you can get

Answer 39

so the absolute max is at (3,4) and that max value is 4

Answer 40

where are you stuck?

Answer 41

I got that now

Answer 42

isnt 0 an abs max too then since its a porabola?

Answer 43

the abs min is 0 because this the lowest point

it happens at (1,0)

Answer 44

Oh got it