Mean Value Theorem: Homework Solution

[Qube]

Homework Statement

http://i.minus.com/jX32eXvLm6FGu.png

Homework Equations

The MVT applies if

1) The function is continuous on the closed interval [a,b] such that a<b.
2) The function is differentiable on the open interval (a,b)

And if the above two conditions are fulfilled then there is some point c between a and b at which the slope is equal to (f(b) - f(a)) / (b-a)

The Attempt at a Solution

1) The function is continuous for all real x. The function has a slope for all real x.
2) The function is differentiable for all x, as stated in the problem.

Therefore the MVT applies.

Because the MVT applies [f(7) - f(1)] / 6 = f'(c).

The maximum that f'(c) can be is 5, as stated in the problem. The slope is always between 2 and 5, including the endpoints. The minimum f'(c) can be is 2.

Therefore the inequality should be 12 ≤ f(7) - f(1) ≤ 30.

 

[LCKurtz]

Homework Statement

http://i.minus.com/jX32eXvLm6FGu.png

Homework Equations

The MVT applies if

1) The function is continuous on the closed interval [a,b] such that a<b.
2) The function is differentiable on the open interval (a,b)

And if the above two conditions are fulfilled then there is some point c between a and b at which the slope is equal to (f(b) - f(a)) / (b-a)

The Attempt at a Solution

1) The function is continuous for all real x. The function has a slope for all real x.
2) The function is differentiable for all x, as stated in the problem.

Therefore the MVT applies.

Because the MVT applies [f(7) - f(1)] / 6 = f'(c).

The maximum that f'(c) can be is 5, as stated in the problem. The slope is always between 2 and 5, including the endpoints. The minimum f'(c) can be is 2.

Therefore the inequality should be 12 ≤ f(7) - f(1) ≤ 30.

I don't see any question. But I do like that last inequality, if you were wondering.

 

[Qube]

Alright. That was what I was looking for. Thank you :)!