Question

Find the point(s) guaranteed by the mean value theorem for f(x)=|x^2-9| over the interval [0,2]

Answer 1

x^2 - 0?

Answer 2

i meant 9 lol

Answer 3

lol

Answer 4

So what are the conditions for the MVT

Answer 5

well it is already guaranteed that the mean value theorem applies so i did f'(x)= f(2)-f(0)/2
f'(x)=-2
then what do i find the derivative of f(x)=|x^2-9| because i dont know how to find the derivative of a absolute value

Answer 6

\[\sqrt{x^2} = |x|\]

Answer 7

how would i do that for |x^2-9| and did i do the beginning of the problem correctly

Answer 8

y =|x^2-9|
Let u =x^2-9
=> \(y=\sqrt{ u^2}\)
\[\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}\]
Does that help?

Answer 9

i still don't understand

Answer 10

Which part?

Answer 11

how i'm supposed to take the derivative of \[|x ^{2}-9|\]

Answer 12

Do you know that \(\sqrt{a^2} = |a| \)?

Answer 13

yes but how do i apply that
\[\sqrt{(x-9)^{2}}\] isn't the derivative

Answer 14

It's the question :P
To find the derivative of |x^2 -9|, you can find the derivative of \(\sqrt{(x^2-9)^2}\), which is the same as |x^2 -9|.

Btw, it should be \(\sqrt{(x^2-9)^2}\), not \(\sqrt{(x-9)^2}\)

Answer 15

oh okkkkkkk but is that even what i'm supposed to be doing in this problem

Answer 16

You need to find f'(x), so, I think so :|

Answer 17

for realsssss noooooooo

Answer 18

Hmmm.. MVT:
Suppose y=f(x) is continuous on a closed interval [a,b] and differentiable on the interior (a,b). Then there is at least one point c at which \(\frac{f(b)-f(a)}{b-a} = f'(c)\)

In this question, the conditions are satisfied, that is f(x) = |x^2-9| is continuous on the closed interval [0, 2] and is differentiable on (0, 2)
So, just apply it: \(\frac{f(2)-f(0)}{2} = f'(c)\)
And you have to find f'(x), isn't it?

Answer 19

yeah

Answer 20

So... just differentiate that guy :|

Answer 21

so the derivative is 2x

Answer 22

x=-1?

Answer 23

not quite

Answer 24

on the interval \([0,2]\) you have \(|x^2-9|=9-x^2\)

Answer 25

how is it 9−x^2

Answer 26

since this is a quadratic, the "point guaranteed to exist by the mean values theorem will be in the midpoint of the interval, namely at \(x=1\)

Answer 27

if \(-3<x<3\) we have \(x^2-9<0\) and so \(|x^2-9|=-(x^2-9)=9-x^2\)

Answer 28

in english \(x^2-9\) is negative on the interval \([0,2]\) and so the absolute value is minus the whole thing , namely \(9-x^2\)

Answer 29

okay so the derivative is -2x right and then you set that equal to the mean and you solve for x and you get x=1?

Answer 30

lets try it

Answer 31

\(f(0)=9\) and \(f(2)=5\) and so \(\frac{f(2)-f(0)}{2-0}=\frac{5-9}{2}=-4\)

Answer 32

isn't it -2

Answer 33

ooops bad arithmetic

Answer 34

oh okay lol

Answer 35

<3

Answer 36

yes it is \(-2\)
now we set \(-2x=-2\) and guess what? \(x=1\)

Answer 37

-2=f'(x)

Answer 38

yessss! alright i get it now

Answer 39

you can convince yourself, if you have the time in inclination, that the "point guaranteed to exist by the mean value theorem" is smack in the middle of the interval if your function is a quadratic