Question

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).

Answer 1

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

Answer 2

Correct assumption.

Answer 3

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./

Answer 4

( first, do the 1st 3 steps )

Answer 5

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

Answer 6

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

Answer 7

21

Answer 8

Ok,
f(0)=1
f(4)=85

slope of the secant = (85-1)/(4-0) = 84/4 = 21

Correct!

Answer 9

Now, find f`(x)

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

yes, that is correct.
f`(x)=3x²+2x+1

Answer 15

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.

Answer 16

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.

Answer 17

what is the answer in the book?

Answer 18

and what answer have you previously been getting?

Answer 19

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

Answer 20

oops, -3 can't be right.

Answer 21

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

Answer 22

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))

Answer 23

you can use the quadratic formula (i would advise)

Answer 24

You would think so, but apparently not.

Answer 25

Ok, should I do this algebraic task for you?

Answer 26

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

Answer 27

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

Answer 28

\(\large\color{black}{ \displaystyle 3x^2+2x+1=21 }\)
\(\large\color{black}{ \displaystyle 3x^2+2x-20=0 }\)

\(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{2^2-4(3)(-20)}}{2\cdot 3} }\)

\(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{224}}{2\cdot 3} }\)

\(\large\color{black}{ \displaystyle \frac{-2\pm \sqrt{61\cdot 4}}{2\cdot 3} }\)

\(\large\color{black}{ \displaystyle \frac{-2\pm 2\sqrt{61}}{2\cdot 3} }\)

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

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

Answer 33

I am using "latex"

Answer 34

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

Answer 35

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

Answer 36

Will do. Thanks.

Answer 37

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

Answer 38

but that is not required I guess....

Answer 39

so, your only answer (that is in the interval [0,4] )

= \(\large\color{black}{ \displaystyle \frac{1 }{3} (\sqrt{61}-1) }\)

Answer 40

any questions ?

Answer 41

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

\(\large\color{black}{ \displaystyle \sqrt{61}\approx 8+\frac{-3}{16} }\)

\(\large\color{black}{ \displaystyle \sqrt{61}\approx 7+\frac{13}{16} }\)

Answer 42

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)

\(\color{black}{ \displaystyle {\rm Approximation~of~f(x)}_{\rm~at~~x~=~a}~~ \\[1.3em] \displaystyle =f(a)+f`(a)(x-a)+\frac{f``(a)}{2!}(x-a)^2+\frac{f```(a)}{3!}(x-a)^2+ \\[1.3em] \displaystyle {\bf ....}~+~+\frac{f^{(n)}(a)}{n!}(x-a)^n }\)

Answer 43

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