Say I have a function
$$g(x) = \int_a^b (f(t)-x)^3dt$$
how would I go about finding the critical points of this function?
I know that FTC gives you that if
$$h(x) = \int_0^x f(t)dt$$ then $$h'(x) = f(x)$$
but I just keep thinking that
$$g(x) = (f(b)-x) – (f(a)-x)$$
so wouldn't the derivative be 0? And since you find critical points where g'(x)=0 wouldn't that mean everywhere is a critical point?
I know this can't be true but having hard time figuring out what to do.
Best Answer
If you want to figure out where $g'$ is zero, compute it as \begin{align*} g'(x) &= -3\int_a^b\, (f(t) - x)^2\, dt \\ &= -3\int_a^b\, f^2(t)\, dt + 6x \int_a^b f(t)\, dt -3x^2(b - a). \end{align*} It's a quadratic in $x$. Set it to zero and solve for the $x$s that are the critical points.
If you set $I_0 = b - a$, $I_1 = \int_a^b\, f(t)\, dt$, $I_2 = \int_a^b\, f^2(t)\, dt$ your critical points occur at \begin{equation*} \hat{x} = \frac{I_1}{I_0} \pm \sqrt{\biggl( \frac{I_1}{I_0}\biggr)^2 - \frac{I_2}{I_0}} \end{equation*}