More than one absolute max?

Question

I have the function $f(x) = 3x - x^3$ evaluating the absolute max and min on the interval $[-2,3]$.

The critical points are $-1$ and $1$. The value of $f(1) = 2$ and the value of $f(-2) = 2$. Are both of these absolute maximums?

EDIT: on the closed interval of $[-2,3]$

Answer 1

Just to confirm your work, if we have $f(x)=3x-x^3$, then $f'(x)=3-3x^2$ which has zeros at $x=\pm 1$. At $x=1$, $f''(1)= -6$ so that $(1,f(1))=(1,2)$ is a local maximum and $f''(-1)=6$ so that $(-1,f(-1))=(-1,-2)$ is a local minimum. If you restrict to the interval $[-2,3]$, then you must also consider the endpoints. Now at $x=-2$, we have $f(-2)=2$ and $f(3)=-18$. Then $f(x)$ has a maximum value $2$ on the interval $[-2,3]$ that happens to occur twice: once at $x=-2$ and once at $x=1$. We can see this function on the interval $[-2,3]$ below:

It is fine for the same maximum for a function to occur many times over a given interval. Take for example $f(\theta)=\sin(2\theta)$ over the interval $[0,2\pi]$. The same maximum and minimum value occur several times, just at different locations. That might be the confusion. While the max/min value occurs multiple times, the points are unique. In your example, the maximum value $2$ occurs twice (so there is still only one absolute maximum, the value $2$) but at it occurs at two distinct points: $(-2,2)$ and $(1,2)$.

Answer 2

Try using the second derivative test to deduce if it is a maximum or a minimum. If the second derivative is negative, then it is concave down a.k.a it is a maximum. Else, it is a minimum.

Answer 3

There are three conflicting concepts.

local max/min which occur when $f'(x_0) =0; f(x_0) = c$. There is absolutely no reason at some other section of the graph (after some inflection points) that there won't be other points where $f(x) = c$ or where $f(x)$ is bigger or less than $c$.

Example $f(1)=2$ is a local maximum and $f(0.9) < 2$ and $f(1.1) < 2$. But $f(-3) = -9 + 27 = 18 > 2$.

global max/min which is the max or min a function ever achieves. This need not actually exist. But if it does it is a unique value. However it may occur and multiple times.

Example: $3x -x^3$ does not have a global maximum/minimum. As $x\to \infty$ $3x - x^3 \to -\infty$ and $x\to -\infty$ $3x - x^3 \to \infty$. $x^2$ has a global minimum at $x_0=0; f(x_0) = 0$ but no global maximum. $x^4 - 2x^2$ has global minimum of $f(-1)=f(1) = -1$ which it achieves twice.

And Absolute max/min are the maximum/minimum value a function achieves within an interval. As the interval is an artificial constraint, this value will is completely determined by the interval.

Now here's the thing! If the function is continuous, then for all the points on which $f'(x)$ does not equal $0$, the function will be either increasing or decreasing. So logistically if absolute max/min will have to happen at either local max/min or at endpoints. (If $x_1$ is so that $f'(x_1)\ne 0$ then $f$ is increasing or decreasing and $x = x_1$ which means to the left of $x_1$ the function will be bigger or smaller and to the right it will be smaller or bigger-- but if $x_1$ is an endpoint... then it is either bigger than all the values to its immediate left or right and we don't have to worry about the values onn the other side.)

So to find absolute maximum on $[-2,3]$ we must find the local max/min and the values of $f(-2)$ and $f(3)$ and compare.

$f'(x) = 3 - 3x^2= 0 \implies x = \pm 1$ so $f(1)= 3 - 1^3 = 2$; $f(-1) = -3+1^3 = -2$ so $f(1)$ is local max, and $f(-1)$ is local min.

$f(-2) = 3(-2) +2^3 = 2$ and $f(3) = 3*3 - 3^3 = -24$.

So we have: $-24 = f(3) < -2 = f(-1) < 2 = f(1) = f(-2)$.

So $f(3) = -24$ is an absolute minimum on the interval (although if we changed the interval this need not stay true. $f(-1) = -2$ is a local minimum but not an absolute minimimal. For all $x > 1$ we have $f(x) < 2$ as well. $f(1) = 2$ is a local maximum. And $f(1) = f(-2) = 2$ is a global maximum; it's the greatest value achieve on this interval but if we expanded the interval any ammount it would no longer be the absolute maximum. For any $x < -2$ we have $f(x) > 2$.

Answer 4

Hi yes in this case both points are considered absolute maxima as an absolute maximum is defined as: A value of a function f on D (domain) if f(c) >= f(x) for all x in D.