Does the Mean Value Theorem Guarantee a Function Value of 4?

[courtrigrad]

If f is continuous and \int^{3}_{1} f(x) dx = 8, show that f takes on the value 4 at least once on the interval [1,3] . I know that the average value of f(x) is 4. So does this imply that f_{ave} = f(c) = 4 and f(x) takes on the value of 4 at least once on the interval [1,3] ?

 

[quasar987]

I'd go with contradiction. Suppose there is no x0 in [1,3] such that f(x0)=4. With that assumption, what does the mean value thm says about the values f can take on [1,3]? There are two cases, etc.

 

[0rthodontist]

If f is continuous and \int^{3}_{1} f(x) dx = 8, show that f takes on the value 4 at least once on the interval [1,3] . I know that the average value of f(x) is 4. So does this imply that f_{ave} = f(c) = 4 and f(x) takes on the value of 4 at least once on the interval [1,3] ?

Yes, but not directly--the mean value theorem is about derivatives. What is the derivative of g(x) = \int_{3}^{x} f(t) dt?

 

[quasar987]

I'd go with contradiction. Suppose there is no x0 in [1,3] such that f(x0)=4. With that assumption, what does the mean value thm says about the values f can take on [1,3]? There are two cases, etc.

make that the intermediate value thm, sorry.

Anyway, courtigrad's way is a little bit faster, more elegant and "on topic" provided this is an exercice designed to make you apply the mean valut thm.

 

[courtrigrad]

Thanks for your help. Let's say I want to prove (not rigorously) the Mean Value Theorem for Integrals, \int_{a}^{b} f(x)\dx = f(c)(b-a) by using the Mean Value Theorem for Derivatives to the function F(x) = \int^{x}_{a} f(t)\dt. We know that F(x) is continuous on [a,x] and differentiable on (a,x) . So F(x) - F(a) = f(c)(x-a) or \frac{\int^{b}_{a} f(t)\dt - \int_{a}^{a} f(t)\dt}{b-a} = f(c)(b-a). So f(c) = \frac{1}{b-a}\int_{a}^{b} f(t)\dt. Is this correct?

Thanks

 

[quasar987]

minus the extra (b-a) in the second to last equation, I'd say this is a perfectly acceptable and rigourous proof of the mean value thm for integral. Nice job!

 

[0rthodontist]

This step
\frac{\int^{b}_{a} f(t)\dt - \int_{a}^{a} f(t)\dt}{b-a} = f(c)(b-a)
is not true. Also instead of just writing equations, you should explain what you're doing--for example you should say, "where c is some value between a and b" and you should say when you use the mean value theorem or fundamental theorem of calculus to justify a step.

Because of the way you've presented it, I'm not sure if you are already aware of this, but what is d(F(x))/dx? What is F(a) and F(b)?

 

[courtrigrad]

\frac{dF(x)}{dx} = f(t). F(a) = 0 and F(b) = \int_{a}^{b} f(t)

 

[quasar987]

(you rock!)

But I agree with ortho on the last part: in an exam, you'd most certainly lose points for not expliciting your reasoning in terms of the condition of applicability of the thm, etc. For instance, the step that goes,

We know that F(x) is continuous on [a,x] and differentiable on (a,x) . So F(x) - F(a) = f(c)(x-a)

should read "We know by one version of the fond thm of calculus that under the assumption of the continuity of f on [a,b], F(x) is continuous on [a,b] , differentiable on (a,b) and that F'(x)=f(x). Therefor, by the mean value thm, there exists a number c in (a,b) such that F(b)-F(a)=F'(c)(b-a). But F'(c) = f(c), F(b)=\int_a^b f(t)dt, F(a)=\int_a^a f(t)dt=0, hence the result."

 

[HallsofIvy]

Let F(x)= \int_1^x f(t)dt. Then F(1)= \int_1^1 f(t)dt= 0 and F(3)= \int_1^3 f(t)dt= 8 so \frac{F(3)- F(1)}{3- 1}= \frac{8- 0}{2}= 4. By the mean value theorem then, there must exist c between 1 and 3 such that F'(c)= f(c)= 4.