Question

If f(x) = ∣(x2 − 6)(x2 + 2)∣, how many numbers in the interval 1 ≤ x ≤ 2 satisfy the conclusion of the mean value theorem?

Answer 1

well first thing's first its really |(x^2-6)(X^2+2)|

Answer 2

also in that interval given how many numbers can apply to the mean value theorem which the formula for it is

Answer 3

you want to prove the mean value theorem saying that

\(\large\color{black}{ \tt f'(c)=\frac{f(a)-f(b)}{a-b} }\).

In other words, to porve that the derivative at x=c, is the slope of the secant between a and b, over the interval, \(\large\color{black}{ \tt [a,b] }\)

So I think you would need to know that it is continuous at this interval \(\large\color{black}{ \tt [1,2] }\).

Well, all I got so far, is that it is a polynomial, and the absolute value is defined
And polynomial is contious on \(\large\color{black}{ \tt (-\infty,\infty) }\)by definiton.

Answer 4

Right. And I thought plugging in the intervals in the equation would solve something but im not positive.

Answer 5

But before anything I take the derivative of both polynomials. Should i do product rule?

Answer 6

\(\Large\color{black}{ \tt f'(c)= \frac{f(2)-f(1)}{2-1} }\) is the slope between the endpoints.

Answer 7

I haven't actually done derivatives of an absolute value though.

Answer 8

I would just distribute and do the power rule.

Answer 9

well i just suppose that it will be a positive number, but absolute values has a cusp, so wouldn't that make the equation undefined at a certain point?

Answer 10

http://www.wolframalpha.com/input/?i=derivative+of+%E2%94%82x%5E4-4x%5E2-12%E2%94%82

Answer 11

did you distribute it already?

Answer 12

Yes I did, now the undefined values.
Yes, lets see.
and
\(\Large\color{black}{ \tt x^4-4x^2-12=0 }\)
\(\Large\color{black}{ \tt x^4-4x^2+4=16 }\)

Let, \(\Large\color{black}{ \tt x^2=a }\)

\(\Large\color{black}{ \tt a^2-4a+4=16 }\)

\(\Large\color{black}{ \tt (a-2)=\pm 4}\)

\(\Large\color{black}{ \tt a=\pm 4+2}\)

\(\Large\color{black}{ \tt a=6,~~-2}\)

Answer 13

\(\Large\color{black}{ \tt x=\pm\sqrt{6},~~~\pm i\sqrt{2}}\)

Answer 14

and the sqrt(6) is greater than 2

Answer 15

where did you get \[x^4-4x^2+4=16\]?

Answer 16

I added 16 to both sides.

Answer 17

where did the 16 come from sorry i got lost.

Answer 18

I wanted to make the left side into a perfect square polynomial.

Answer 19

(which I did)

Answer 20

okay got it

Answer 21

but, you want to set the derivative equal to the slope of the secant.

Answer 22

tell me what is:
f(2)=?
f(1)=?

Answer 23

f(2)=12
f(1)=15

Answer 24

I think

Answer 25

By the way the answer choices I have if it is not to late is

Zero
One
Two
Three

Answer 26

connection snapped, but f(2) and f(1) are correct.

Answer 27

Alright great so far.

Answer 28

\(\LARGE\color{black}{ \tt f'(c) = \frac{ f(2) - f(1) }{ 2 - 1 } }\)

plug in your values. And here you can use this,
copy the gray and use it....
`\(\LARGE\color{black}{ \tt f'(c) = \frac{ f(2) - f(1) }{ 2 - 1 } }\)`

Answer 29

(\tt s just the font)

Answer 30

so let's see

Answer 31

fill in your values for f(2) and f(1)... and simplify.

Answer 32

\[12-15/2-1\]

Answer 33

Use my thing in gray copy it, and go for it...

Answer 34

and i'd get \[-3\]

Answer 35

Yes.

Answer 36

So, now, set the derivative equal to that.

Answer 37

okay so what did we say the derivative was?

Answer 38

I was thrown off by it.

Answer 39

\(\LARGE\color{black}{ \tt f'(x)=\frac{(x^4-4x^2-12)(4x^3-8x)}{|4x^2-4x^2-12|} }\)

Answer 40

\(\LARGE\color{black}{ \tt -3=\frac{(x^4-4x^2-12)(4x^3-8x)}{|4x^2-4x^2-12|} }\)

Answer 41

solve for x....

Answer 42

i got 84

Answer 43

wait.

Answer 44

How do you get only 1 solution?

Answer 45

wasn't i supposed to plug it in?

Answer 46

I have a question, @freckles , can we cancel the \(\LARGE\color{black}{ \tt 4x^2-4x^2-12 }\) with the \(\LARGE\color{black}{ \tt |4x^2-4x^2-12| }\) ?

Answer 47

I mean since we are only dealing with positive xs? and in this case, not in general.

Answer 48

if 4x^2-4x^2-12>0 then |4x^2-4x^2-12|=4x^2-4x^2-12
if 4x^2-4x^2-12<0 then |4x^2-4x^2-12|=-(4x^2-4x^2-12)
so it would depend on the restrictions of x and if that restriction is a subset of the first inequality or the second inequality I mentioned

Answer 49

\(\LARGE\color{black}{ \tt f'(x)=\frac{(x^4-4x^2-12)(4x^3-8x)}{|4x^4-4x^2-12|} }\)
Correction.

Answer 50

http://www.wolframalpha.com/input/?i=x%5E4-4x%5E2-12%3E0
x>sqrt(6)
x<-sqrt(6)

So in this case, where our interval is 2 and 1,. we would need to multiply the top
x^4-4x^2-12 times -1, and then cancel it with the absolute value.

I mean like

\(\LARGE\color{black}{ \tt f'(x)=\frac{(x^4-4x^2-12)(4x^3-8x)}{|4x^4-4x^2-12|} }\)

\(\LARGE\color{black}{ \tt f'(x)=\frac{-(x^4-4x^2-12)(8x-4x^3)}{|4x^4-4x^2-12|} }\)

\(\LARGE\color{black}{ \tt f'(x)=\frac{\bcancel{\cancel{\color{red}{-(x^4-4x^2-12)}}}(8x-4x^3)}{\bcancel{\cancel{\color{red}{|4x^4-4x^2-12|}}} } }\)

Answer 51

right?

Answer 52

But how would the x^4 and 4x^4 cancel?

Answer 53

If f(x) = ∣(x2 − 6)(x2 + 2)∣, how many numbers in the interval 1 ≤ x ≤ 2 satisfy the conclusion of the mean value theorem?


f(x)=(x^2+2)|x^2-6|
x^2+2>0 for all real x
f is not differentiable at sqrt(6) or -sqrt(6)
g(x)=x^2-6 looks like:

it is negative between -sqrt(6) and +sqrt(6)
so that is where fbi got the - sign
so since our function is restricted on 1 to 2 we have
f(x)=-(x^2+2)(x^2-6)
differentiate that by product rule or multiply it all out
then differentiate
f(x)=-(x^4-4x^2-12)
f'(x)=-4x^3+8x on the interval (1,2)