Find a function increasing on the interval $[0,1]$ whose average is $\frac23$?

Question

So I am not sure how to go about this. I've been using the average value theorem to calculate the average value, but how do I work backwards to find the function and am given the average value itself?

Answer 1

What is usually good in a situation like this is to come up with a family of functions depending on a parameter. What if $f(x)=x+a$ for some $a\in\mathbb R$? What if $f(x)=x^b$ for some $b>0$? If you don't have an idea at first, take a function depending on a parameter and compute the average. Now, can you choose the parameter so that you get what you want?

Answer 2

Pick $f(x) = cx$

We want: $\int_0^1cxdx = \frac{2}{3}$ for some $c$

$$\int_0^1cxdx = c\left [\frac{x^2}{2} \right ]_0^1 = \frac{c}{2}$$

$$\implies c = \frac{4}{3}$$

therefore, $$f(x) = \frac{4}{3}x$$

is increasing on $[0, 1]$ and has average value $\frac {2}{3}$.

Answer 3

Let $f(x)=ax$, $a>0$, $$\frac23 = \frac1{1-0}\int_0^1 ax \, dx$$

Now, you can solve for $a$.

Note that in general, the solution is not unique, in fact, we can let $f(x)=ag(x)$ be any positive increasing function and solve for $a$.