Mean Value Theorem

[stinajeana]

Find every value of c that satisfies the conclusion of the Mean Value Theorem for f(x)=(4x^2)/(x-1) on the interval [2,5]

Did I do this right?

f(x)=(4x^2)/(x-1)
=f'(x)=(x-1)*(8x)-(4x^2)*(1)
=f'(x)=(8x^2)-(8x)-(4x^2)
=f'(x)=4x(x-2)

f'(c)=f(b)-f(a)/b-a
f'(c)=f(5)-f(2)/5-2
f'(c)=25-16/3
f'(c)=9/3
f'(c)=3

4x=3
x=3/4
x=.75

x-2=3
x=3+2
x=5

....so because 5 is in between the interval [2,5] c=5

 

[stinajeana]

Nevermind... I just realized what I did wrong. I forgot the (x)^2 from the quotient rule

 

[mmm4444bot]

Find every value of c that satisfies the conclusion of the Mean Value Theorem for f(x)=(4x^2)/(x-1) on the interval [2,5]


Did I do this right?


f'(c)=3

Yes, 3 is correct.

4x=3
x=3/4
x=.75


x-2=3
x=3+2
x=5

This part looks incorrect. Setting those factors from f'(x) equal to 3 is not the way to go, for finding c.

Now that you've remembered the (x-1)^2 part, in the derivative of function f, you know that f'(c) = 4c(c-2)/(c-1)^2, yes?

So that is what equals 3. Solve for c.

Cheers :cool:

 

[stinajeana]

Yes, 3 is correct.







This part looks incorrect. Setting those factors from f'(x) equal to 3 is not the way to go, for finding c.

Now that you've remembered the (x-1)^2 part, in the derivative of function f, you know that f'(c) = 4c(c-2)/(c-1)^2, yes?

So that is what equals 3. Solve for c.

Cheers :cool:

Yes, I got -1 and 3 but c=3 because -1 is not between 2 and 5. Is this correct?

 

[mmm4444bot]

Correct. :cool:

 

[stinajeana]

Correct. :cool:

If its not too much... would you mind showing me how you did it? because I found the derivative, simplified, plugged a and b into the formula and got 3 then took the simplified derivative and made it equal to 3 and then solved for x. I'm really paranoid that I did a step wrong

 

[mmm4444bot]

I did it the same way as you just described.

Here is a graph of function f, shown in red. The line connecting (2,16) to (5,25) is shown in green; it has slope 3 -- from your calculation (25-16)/(5-2), and so c=3.

The theorem says that somewhere along the interval [2,5], there is a line tangent to f which also has slope c (that is, the derivative of f is 3, somewhere in the interval). That tangent line is parallel to the green line because their slopes are the same.

You found that location: x=3.

The tangent line to f at (3,18) is shown in blue.

 

[stinajeana]

I did it the same way as you just described.

Here is a graph of function f, shown in red. The line connecting (2,16) to (5,25) is shown in green; it has slope 3 -- from your calculation (25-16)/(5-2), and so c=3.

The theorem says that somewhere along the interval [2,5], there is a line tangent to f which also has slope c (that is, the derivative of f is 3, somewhere in the interval). That tangent line is parallel to the green line because their slopes are the same.

You found that location: x=3.

The tangent line to f at (3,18) is shown in blue.

View attachment 2939

Thank you!