anonymous
  • 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).
Calculus1
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
I am going to assume you know the Mean Value Theorem. Okay?
anonymous
  • anonymous
Correct assumption.
SolomonZelman
  • 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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SolomonZelman
  • SolomonZelman
( first, do the 1st 3 steps )
anonymous
  • anonymous
I did all that. Somehow I'm messing up.
SolomonZelman
  • SolomonZelman
ok, what was your slope of the secant from x=0 to x=4?
anonymous
  • anonymous
21
SolomonZelman
  • SolomonZelman
Ok, f(0)=1 f(4)=85 slope of the secant = (85-1)/(4-0) = 84/4 = 21 Correct!
SolomonZelman
  • SolomonZelman
Now, find f`(x)
SolomonZelman
  • 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
  • anonymous
I've done that twice. I must be making an error.
SolomonZelman
  • SolomonZelman
you have a problem taking the derivative of ` x³+x²+x+1 ` , is that correct ?
anonymous
  • anonymous
No. I get 3x^2 + 2x +1
SolomonZelman
  • SolomonZelman
yes, that is correct. f`(x)=3x²+2x+1
SolomonZelman
  • 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
  • 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
  • SolomonZelman
what is the answer in the book?
SolomonZelman
  • SolomonZelman
and what answer have you previously been getting?
anonymous
  • anonymous
I got x = -3 and x = 7/3. The book says it's (sqrt(61 - 1)/3.
anonymous
  • anonymous
oops, -3 can't be right.
SolomonZelman
  • SolomonZelman
yes, the book is correct.... :) (well, duh, but I actually did it...)
SolomonZelman
  • 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
  • SolomonZelman
you can use the quadratic formula (i would advise)
anonymous
  • anonymous
You would think so, but apparently not.
SolomonZelman
  • SolomonZelman
Ok, should I do this algebraic task for you?
SolomonZelman
  • SolomonZelman
well, we are getting the calculus algorithm, so that wouldn't do much harm....
anonymous
  • anonymous
Well, that seems to be where my brain is malfunctioning.
SolomonZelman
  • 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
  • SolomonZelman
from here you can proceed with no probs....
anonymous
  • 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
  • SolomonZelman
oh, it is just a minor mistake... messes everything up but happens to everybody:)
anonymous
  • anonymous
How do you get the math characters in the display like that?
SolomonZelman
  • SolomonZelman
I am using "latex"
anonymous
  • anonymous
OK, thanks. I'll have to research that some time. Thanks again.
SolomonZelman
  • SolomonZelman
you can look up a "legendary latex tutorial" online, and you should find a tutorial made by thomaster
anonymous
  • anonymous
Will do. Thanks.
SolomonZelman
  • SolomonZelman
If you want, we can do a polynomial approximation of √61, using f(x)=√x, for x=64.
SolomonZelman
  • SolomonZelman
but that is not required I guess....
SolomonZelman
  • SolomonZelman
so, your only answer (that is in the interval [0,4] ) = \(\large\color{black}{ \displaystyle \frac{1 }{3} (\sqrt{61}-1) }\)
SolomonZelman
  • SolomonZelman
any questions ?
SolomonZelman
  • 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
  • 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
  • SolomonZelman
in any case... won't overwhelm more than what I did already

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