The maximum value of a function

Question

Is it a general rule/theorem that the abscissa of the maximum value of a function is where the derivative of that function is equal to zero ? I was asked to find the abscissa of when an integral is maximum, so I figured it would be the same abscissa of where its derivative is equal to zero. Thanks in advance.

Answer 1

It's not possible for a function to have an extremum (maximum or minimum) if the derivative is equal to something other than zero. Note that if the derivative is undefined, then the derivative is not equal to something other than zero. (Some care is necessary when considering endpoints.)

To see why this is, suppose you're looking for the maximum value, and the derivative is positive. Then you could increase the value of the function by increasing the abscissa (for those unfamiliar with "abscissa", it basically means "x"). If the derivative is negative, then you could find a larger value by decreasing the abscissa.

For an integral, integration is the inverse of differentiation (to within a constant, which is why you need the +C for indefinite integrals). So the derivative of the integral of a function is just that function. If the function is well-defined everywhere, then the maximum will occur either at an end point (if you are restricted to an interval), or where the function is zero.

Answer 2

There are some qualifications. The function must be differentiable, and it must have a maximum in the interior of the interval in question (not at an endpoint). If so, the derivative of the function at the maximum is $0$.

There may be more than one point where the derivative is $0$: they may be local maxima, or local minima, or points of inflection.

Answer 3

Yes, there is a theorem that states this, look up optimality conditions for unconstrained optimization. What you stated is the first-order necessary optimality condition. Here's a link from the first hit in a google search, its the first theorem. You can also check a 1D or 3D Calculus textbook.

If it's of the following form $F(t) = \int^t_0 f(\tau) \mathrm{d}\tau$, the you can take the derivative (via the fundamental theorem of calculus) $$F'(t) = f(t)$$ and set the derivative equal to zero to find critical values. If the integral has constant limits of integration, e.g. $F(t) = \int^{t_1}_{t_0} f(\tau) \mathrm{d}\tau$ then $F'(t) = 0$.

Answer 4

To illustrate the answer of Robert Israel, as an example of a function not differentiable at the local maximum (and that can be defined by an integral) consider: $$ f(x)=\frac{3}{2}(1-\sqrt[3]{x^2})=\int_{-1}^x\frac{-1}{\sqrt[3]{t}}dt $$

That has a maximum at $x=0$ where it is not differentiable.