Question

Mean Value Theorem Question. Please help!

Answer 1

\leq

Answer 2

the suspense is killing me

Answer 3

Suppose that \[3 lef'(x) \le5 \]for all values of x on the real number line. Use the mean value theorem to show that \[18lef(8)-f(2)\le30\]

Answer 4

sorryy, I'm still trying to get the hang using the equation symbols and stuff.

Answer 5

\[3\leq f'(x)\leq 5\]
\[15\leq f(8)-f(2)\leq ?\]

Answer 6

I still failed miserably at it.

Answer 7

Bingo!

Answer 8

right click and you can see the code

Answer 9

the first line \(3\leq f'(x)\leq 5\) says that \(f\) cannot grow any slower than a line with a slope of 3, or any faster than a line with a slope of 5

Answer 10

Okay .

Answer 11

i am not sure what the last line of the question is though

Answer 12

are you supposed to fill in the question mark here
\[15\leq f(8)-f(2)\leq ?\]??

Answer 13

30

Answer 14

ok you are told
\[15\leq f(8)-f(2)\leq 30\] right? then what is the question?

Answer 15

Oh, and it's 18≤f(8)−f(2)≤30, not 15≤f(8)−f(2)≤30

Answer 16

Sorry, hang on.

Answer 17

\[18\leq f(8)-f(2)\leq 30\]

Answer 18

Suppose that 3≤f′(x)≤5 for all values of X on the real number line. Use the Mean Value Theorem to show that 18≤f(8)−f(2)≤30.

Answer 19

That's the entire question.

Answer 20

I tried doing it, and this is what I got:
f(8) - f(2) = f'(c) (8-0)
f(8) = f(2) + 8f'(c)
f(8) = 30 + 8(5)
f(8) = 70

Answer 21

Doubt that's correct though.

Answer 22

ok we know that \(\frac{f(8)-f(2)}{8-2}=\frac{f(8)-f(2)}{6}=f'(c) \) for some \(c\in (2,8)\)

Answer 23

and we know that \(3\leq f'(c)\leq 5\)

Answer 24

Okay

Answer 25

that pretty much does it, right?

Answer 26

we get
\[3\leq \frac{f(8)-f(2)}{6}\leq 5\implies 18\leq f(8)-f(2)\leq 30\] by multiplying by 6

Answer 27

That's it? We don't need to plug in any numbers or anything?

Answer 28

no it is an inequality
that is the whole story

Answer 29

okay, thank you! :)

Answer 30

yw