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.