- anonymous

If f(x)= x^3 + x^2 + x +1, find a number c that satisfies the conclusion of the Mean Value Theorem on the interval [0,4]. (This seems very straight forward, but I keep getting the wrong answer).

- jamiebookeater

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- SolomonZelman

I am going to assume you know the Mean Value Theorem. Okay?

- anonymous

Correct assumption.

- SolomonZelman

You need to:
1) Find f(0)
2) Find f(4)
3) Find the average rate of change betwen (0,f(0)) and (4,f(4))
then we will find point(s) that have the same slope as the one you find in step 3./

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## More answers

- SolomonZelman

( first, do the 1st 3 steps )

- anonymous

I did all that. Somehow I'm messing up.

- SolomonZelman

ok, what was your slope of the secant from x=0 to x=4?

- anonymous

21

- SolomonZelman

Ok,
f(0)=1
f(4)=85
slope of the secant = (85-1)/(4-0) = 84/4 = 21
Correct!

- SolomonZelman

Now, find f`(x)

- SolomonZelman

And then set the f`(x)=c, in this case 21, and we will see at which points does the function have an instanteneous slope of 21.

- anonymous

I've done that twice. I must be making an error.

- SolomonZelman

you have a problem taking the derivative of ` x³+x²+x+1 ` , is that correct ?

- anonymous

No. I get 3x^2 + 2x +1

- SolomonZelman

yes, that is correct.
f`(x)=3x²+2x+1

- SolomonZelman

REMEMBER! The derivative of the function, is its (the function's) slope.
Now, the slope of the secant in our case is 21. And this is why you need to set f`(x)=21, to see at which points will the derivative of the function (or the function's slope) be =21.

- anonymous

Yes, I understand that. That's what I did. I think I must be making some stupid mistake, because I'm getting a different answer than the book.

- SolomonZelman

what is the answer in the book?

- SolomonZelman

and what answer have you previously been getting?

- anonymous

I got x = -3 and x = 7/3. The book says it's (sqrt(61 - 1)/3.

- anonymous

oops, -3 can't be right.

- SolomonZelman

yes, the book is correct.... :) (well, duh, but I actually did it...)

- SolomonZelman

when you set the equation 3x²+2x+1=21, you will get two solutions, and you will exclude one of them because this another solution is NOT on the interval of [0,4].
Can you solve 3x²+2x+1=21 ? (Shouldn't be a problem for you))

- SolomonZelman

you can use the quadratic formula (i would advise)

- anonymous

You would think so, but apparently not.

- SolomonZelman

Ok, should I do this algebraic task for you?

- SolomonZelman

well, we are getting the calculus algorithm, so that wouldn't do much harm....

- anonymous

Well, that seems to be where my brain is malfunctioning.

- SolomonZelman

\(\large\color{black}{ \displaystyle 3x^2+2x+1=21 }\)
\(\large\color{black}{ \displaystyle 3x^2+2x-20=0 }\)
\(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{2^2-4(3)(-20)}}{2\cdot 3} }\)
\(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{224}}{2\cdot 3} }\)
\(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{61\cdot 4}}{2\cdot 3} }\)
\(\large\color{black}{ \displaystyle \frac{-2\pm 2\sqrt{61}}{2\cdot 3} }\)

- SolomonZelman

from here you can proceed with no probs....

- anonymous

OMG! Dyslexia strikes again. Instead of putting in 2^2, i put in 4 AND THEN squared that. Thanks so much!! This kind of thing makes me crazy.

- SolomonZelman

oh, it is just a minor mistake... messes everything up but happens to everybody:)

- anonymous

How do you get the math characters in the display like that?

- SolomonZelman

I am using "latex"

- anonymous

OK, thanks. I'll have to research that some time. Thanks again.

- SolomonZelman

you can look up a "legendary latex tutorial" online, and you should find a tutorial made by thomaster

- anonymous

Will do. Thanks.

- SolomonZelman

If you want, we can do a polynomial approximation of √61, using f(x)=√x, for x=64.

- SolomonZelman

but that is not required I guess....

- SolomonZelman

so, your only answer (that is in the interval [0,4] )
= \(\large\color{black}{ \displaystyle \frac{1 }{3} (\sqrt{61}-1) }\)

- SolomonZelman

any questions ?

- SolomonZelman

So (extra part) you have a function: \(\large\color{black}{ \displaystyle f(x)=\sqrt{x} }\)
\(\large\color{black}{ \displaystyle L(x)=f(a)+f`(a)(x-a) }\)
this is a linear approximation of the function f(x) near the a point x=a.
in this case our function is f(x)=√x, and a=64.
(our f`(x) is going to be 1/(2√x), and f`(a)=f`(8)=1/(2√64) ...)
\(\large\color{black}{ \displaystyle L(x)=\sqrt{64}+\frac{1}{2\sqrt{64}}(x-64) }\)
\(\large\color{black}{ \displaystyle L(x)=8+\frac{1}{16}(x-64) }\)
this is the tangent line to √x, at x=64
now, you can plug in 61 to get an approzimate value for √61
\(\large\color{black}{ \displaystyle \sqrt{61}\approx 8+\frac{-3}{16} }\)
\(\large\color{black}{ \displaystyle \sqrt{61}\approx 7+\frac{13}{16} }\)

- SolomonZelman

But, by taylor approximation, you would have a closer approximation. (that is, drawing a tangent the the f(x) polynomial of nth degree thorugh the point x=a)
\(\color{black}{ \displaystyle {\rm Approximation~of~f(x)}_{\rm~at~~x~=~a}~~ \\[1.3em] \displaystyle =f(a)+f`(a)(x-a)+\frac{f``(a)}{2!}(x-a)^2+\frac{f```(a)}{3!}(x-a)^2+ \\[1.3em] \displaystyle {\bf ....}~+~+\frac{f^{(n)}(a)}{n!}(x-a)^n }\)

- SolomonZelman

in any case... won't overwhelm more than what I did already

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