Mean Value Theorem: Suppose f is cont. on [3, 7] and -5 <

[chucknorrisfish]

Suppose f(x) is continuous on the interval [3, 7] and -5 < f'(x) < 4 for all x in the interval (3, 7).

Use the Mean Value Theorem to estimate f(7) - f(3).

I know the MVT uses [(b) - f(a)] / [b - a] on the interval [a, b], but I'm not sure how to use that with this question.

 

[stapel]

The Mean Value Theorem says, specifically, that, for f(x) differentiable on an interval (a, b) and continuous on [a, b], the following holds:

. . . . .\(\displaystyle \L f'(c)\, =\, \frac{f(b)\, -\, f(a)}{b\, -\, a}\)

...for at least one value x = c within the interval (a, b). That is, for some a < c < b, the above must be true. (There may be many such c-values, but you are guaranteed of one.)

You are told that f'(x) is between -5 and 4. Then:

. . . . .\(\displaystyle \L -5\, \leq\, \frac{f(b)\, -\, f(a)}{b\, -\, a}\, \leq \, 4\)

You know the values for a and b. Plug them in. Then solve for the difference, and note the inequalities which provide the bounds -- that is, the estimate -- of the difference.

Eliz.

 

[chucknorrisfish]

but then all i would have is -5 <= f(7) - f(3)/ 4 <= 4 right?

 

[stapel]

And then multiply through by 4....

Eliz.