Need help with the mean value theorem

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?

Question options:

1)	None
2)	One
3)	Two
4)	Three

Answer 1

Let consider function g(x) = (x² - 6)(x² + 2). Now g(x) < 0, (x² - 6)(x² + 2) < 0 equivalent x² - 6 < 0, because

x² + 2 alwais positive;.- sqrt(6) < x < sqrt(6). Our interval [1,2] completely inside interval where g(x) < 0.

ecause f(x) = ⌊g(x)⌊ then f(x) = - (x² - 6)(x² + 2) = (6 - x²)(x² + 2).

It satisfy mean value theorem on interval [1, 2]:

1. f(x) continious on [1, 2] as polynomial;
2. f'(x) exist on (1, 2);

Then f'(c) = (f(2) - f(1))/(2 - 1) = 12 - 15 = - 3

f(x) = - x⁴ + 4x² + 12 , f'(x) = - 4x³ + 8x; we get equation for c: - 4c³ + 8c + 3 = 0, 4c³ - 8c - 3 = 0,

This polynomial function h(c) = 4c³ - 8c - 3 has exactly one positive zero (becouse one change of sign). It exactly between 1 and 2 by intermediate value theorem(h(1) = -7< 0, h(2) = 13 > 0).

Because on interval [1, 2] f(x) has only 1 number sutisfi\ying conclusion of mean value theorem.

Answer 2

Hi Marietta T.,

Ordinarily I don't respond to multiple questions from a student asked on the same day. But this and the previous one (well, probably asked later, in fact) are so sufficiently different, that I will.

Mean Value Theorem says that there is at least 1 point on the interval for which the slope of the function at that point equals the average slope of the function across the interval.

I don't have the intuition to guess how many points that might be, so let's just crank and compare.

f(2) = |(4-6)*(4+2)| = |(-2)*(6)| = |-12| = 12

f(1) = |(1-6)*(1+2)| = |(-5)*(3)| = |-15| = 15

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

[f(2)-f(1)]/[2-1] = -3

OK, now we need to crank on the actual function a bit, figure the slope function values.

First off, notice that that the function inside the || stays negative over the entire interval -- therefore, the absolute value operation flips it signwise over the entire interval. Therefore, we may calculate the value of the function inside the ||, and its derivatives, etc. and then flip each result signwise.

OK, so what is f ' [(x^2-6)*(x^2+2)] ? Use the Chain Rule:

= (x^2-6)*2x + 2x(x^2+2) =4x^3 - 8x

and we need this = 3 to satisfy the Mean Value Theorem. Now, there is a closed-form solution for a cubic equation such as this, but really, why not just throw the expression "4x^3 - 8x - 3" onto a spreadsheet, and crank through 1-> 2 interval by 0.01 increments? See what you find!

-- Cheers, -- Mr. d.

Answer 3

Graph this function.

You will immediately see that there is only one point where the slope of the secant is the same as the slope of the function.