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anonymous
 one year ago
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).
anonymous
 one year ago
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).

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4I am going to assume you know the Mean Value Theorem. Okay?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4You 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./

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4( first, do the 1st 3 steps )

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did all that. Somehow I'm messing up.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4ok, what was your slope of the secant from x=0 to x=4?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4Ok, f(0)=1 f(4)=85 slope of the secant = (851)/(40) = 84/4 = 21 Correct!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4And 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
 one year ago
Best ResponseYou've already chosen the best response.0I've done that twice. I must be making an error.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4you have a problem taking the derivative of ` x³+x²+x+1 ` , is that correct ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. I get 3x^2 + 2x +1

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4yes, that is correct. f`(x)=3x²+2x+1

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4REMEMBER! 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.4what is the answer in the book?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4and what answer have you previously been getting?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got x = 3 and x = 7/3. The book says it's (sqrt(61  1)/3.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, 3 can't be right.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4yes, the book is correct.... :) (well, duh, but I actually did it...)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4when 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
 one year ago
Best ResponseYou've already chosen the best response.4you can use the quadratic formula (i would advise)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You would think so, but apparently not.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4Ok, should I do this algebraic task for you?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4well, we are getting the calculus algorithm, so that wouldn't do much harm....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, that seems to be where my brain is malfunctioning.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4\(\large\color{black}{ \displaystyle 3x^2+2x+1=21 }\) \(\large\color{black}{ \displaystyle 3x^2+2x20=0 }\) \(\large\color{black}{ \displaystyle \frac{2\pm \sqrt{2^24(3)(20)}}{2\cdot 3} }\) \(\large\color{black}{ \displaystyle \frac{2\pm \sqrt{224}}{2\cdot 3} }\) \(\large\color{black}{ \displaystyle \frac{2\pm \sqrt{61\cdot 4}}{2\cdot 3} }\) \(\large\color{black}{ \displaystyle \frac{2\pm 2\sqrt{61}}{2\cdot 3} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4from here you can proceed with no probs....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OMG! 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
 one year ago
Best ResponseYou've already chosen the best response.4oh, it is just a minor mistake... messes everything up but happens to everybody:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do you get the math characters in the display like that?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4I am using "latex"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OK, thanks. I'll have to research that some time. Thanks again.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4you can look up a "legendary latex tutorial" online, and you should find a tutorial made by thomaster

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4If you want, we can do a polynomial approximation of √61, using f(x)=√x, for x=64.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4but that is not required I guess....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4so, your only answer (that is in the interval [0,4] ) = \(\large\color{black}{ \displaystyle \frac{1 }{3} (\sqrt{61}1) }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4So (extra part) you have a function: \(\large\color{black}{ \displaystyle f(x)=\sqrt{x} }\) \(\large\color{black}{ \displaystyle L(x)=f(a)+f`(a)(xa) }\) 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) ...) \(\large\color{black}{ \displaystyle L(x)=\sqrt{64}+\frac{1}{2\sqrt{64}}(x64) }\) \(\large\color{black}{ \displaystyle L(x)=8+\frac{1}{16}(x64) }\) 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
 one year ago
Best ResponseYou've already chosen the best response.4But, 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) \(\color{black}{ \displaystyle {\rm Approximation~of~f(x)}_{\rm~at~~x~=~a}~~ \\[1.3em] \displaystyle =f(a)+f`(a)(xa)+\frac{f``(a)}{2!}(xa)^2+\frac{f```(a)}{3!}(xa)^2+ \\[1.3em] \displaystyle {\bf ....}~+~+\frac{f^{(n)}(a)}{n!}(xa)^n }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.4in any case... won't overwhelm more than what I did already
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