Hi,
Here's the problem:
If f(x) = x³ - x² - 4x + 2 and 0 ≤ x ≤ 3. For what value of c is the conclusion of the mean value theorem true?
The answer is:
(1 + sqrt19) / 3
If I could get a detailed explanation, I'd really appreciate it. Thanks.
Hi,
Here's the problem:
If f(x) = x³ - x² - 4x + 2 and 0 ≤ x ≤ 3. For what value of c is the conclusion of the mean value theorem true?
The answer is:
(1 + sqrt19) / 3
If I could get a detailed explanation, I'd really appreciate it. Thanks.
hmm.. maybe i can try answering..
Mean value theorem states that there exist a point c such that the gradient at point c is the same as the gradient of the line when the two end points are connected..
So maybe first you can find
f(0) and f(3)..
Then u find the gradient of the line connecting the 2 points..
then you find the expression for f'(c)..
f'(c) should be equals to the gradient of the line connecting the two end point..
then find c...
Hope it helps..
f(0) = 2
f(3) = 8
When you say gradient, are you talking about slope? If so, s(t2) - s(t1) / t2 - t1.
From that, I got (8 - 2)/(3 - 0), which gives me 6 / 3, or 2. Is this f'(c)? And if so, what do I do from this point to find c? I don't quite understand.
The gradient is the slope. But the slope is the derivitive.
So $\displaystyle f'(x)=\frac{d f(x)}{dx}$. Then you know that when x=c, this new function will be the same as the gradient of the line between the two endpoints. You know what the gradient (slope) of the line between the two endpoints is. Solve for c.
You will get a polynomial, so remember if c is not between 0 and 3 you are not interested in it.
As an aside to others reading this, is it called the gradient because that is valid in more than one dimension? End aside.
I absolutely have problems with my algebra. Unfortunately, I still need to figure out how to do this problem. I'm not trying to give you a hard time. I just need to understand. I understand the derivative part of the polynomial, but where did the 2 come from? Are you using 2 because it falls in the domain 0 < x < 3?
You must not undertsand what you're doing, there's no point trudging through it unless you have some understanding.
Created with Camtasia Studio 5
Also, here are a lot of helpful calc videos to clear up any questions you might have
Larry Green's Calculus Videos
Once, again, before you can do this problem you need to know what the "mean value theorem" says!
The whole point is to find c such that [tex]\frac{f(3)- f(0)}{3- 0}= f'(c).
$\displaystyle f(3)= 3^3- 3^2- 4(3)+ 2$$\displaystyle = \frac{27- 9- 12+ 2= 8$ and $\displaystyle f(0)= 0^3- 0^2+ 4(0)+ 2= 2$ so f(3)- f(0)= 8- 2= 6.
$\displaystyle \frac{f(3)- f(0)}{3- 0}= \frac{6}{3}= 2$.
That's where Plato got the "2"!
Since $\displaystyle f'(x)= 3x^2- 2x- 4$ you want to find c such that
$\displaystyle f'(c)= 3c^2- 2c- 4= 2$. Solve the quadratic equation $\displaystyle 3c^2- 2c- 6= 0$.