Let \(f\) be a function. Consider the function \(A(t) = \frac {1}{t} \int _0^t f(x)\d x \), which gives the average of \(f\) on the interval \([0,t]\), for \(t>0\).
Suppose that \(A\) has a local maximum at \(t=b\), then \(A'(b) = \answer {0}\) (answer with a number). Now, the formula for \(A'(t)\) is
\begin{align*} A'(t) &= \answer {\frac {f(t)}{t}} - \answer {\frac {1}{t^2}}\int _0^tf(x)d x\\ &= \frac {1}{t}\left (\answer {f(t)}-A(t)\right ) \end{align*}
Now since \(A'(b) = \answer {0}\) (answer with a number), we can conclude that \(A(b) = \answer {f(b)}\) (answer with an expression in terms of \(f\)).
Suppose \(f(x) = x(1-x)\). For what value of \(t\) is \(A(t)\) maximized?
You have to find a critical point of \(A\).
This means that we have to solve the equation
\[ A'(x)=0. \]
That is equivalent to solving \[ f(x)=A(x). \]
let’s solve the equation
\[ f(x)=A(x). \]
\[ x(1-x)=\frac {1}{x} \int _0^x f(t)\d t \]
\[ x-x^2=\frac {1}{x} \int _0^x (t-t^2)\d t \]
\[ x-x^2= \frac {x}{2}-\frac {x^2}{3} \]
Divide by \(x\). \[ 1-x= \frac {1}{2}-\frac {x}{3} \]
Solve for
\(x\). How do we know that the function \(A\) has a maximum at this point? \[ t = \answer {\frac {3}{4}} \]