8.2 Finding global maxima and minima

One way to avoid a global minimum is to have $\ds\lim_{x \to \infty}f(x)=-\infty$. Since $f(x)$ keeps getting lower and lower, there is no one value that is the lowest.

Two examples are given below, but many are possible.

Two examples are given below, but many are possible.

If $f(x)=-x^2$ or $f(x)=-\sqrt{|x|}$, then $f(x)$ has a global maximum at $x=0$. Since $f(x)$ keeps getting more and more strongly negative as $x$ gets farther and farther from 0, $f(x)$ has no global minimum.

Try allowing the function to approach the $x$-axis without ever touching it.

Two examples are given below, but many are possible.

Two examples are given below, but many are possible.

If $f(x)=e^x$, then $f(x) > 0$ for all $x$. As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching 0 but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum. Likewise, $f(x)$ increases more and more as we move right, so there is no maximum.

If $f(x)=\arctan(x)+2$, then $f(x) >\left(-\frac{\pi}{2}\right)+2>0$ for all $x$.

As we move left along the $x$-axis, $f(x)$ gets smaller and smaller, approaching $\left(-\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets smaller and smaller as we move left, there is no global minimum.

Likewise, as we move right along the $x$-axis, $f(x)$ gets bigger and bigger, approaching $\left(\frac{\pi}{2}+2\right)$ but never reaching it. Since $f(x)$ gets bigger and bigger as we move right, there is no global maximum.

Since the global minimum value occurs at $x=5$ and $x=-5$, it must be true that $f(5)=f(-5)$.

One possible answer:

Since $f(5)$ is a global minimum, $f(5) \leq f(x)$ for all $x$, and so in particular $f(5) \leq f(-5)$.
Similarly, $f(-5) \leq f(x)$ for all $x$, so in particular $f(-5) \leq f(5)$.
Since $f(-5) \leq f(5)$ AND $f(5) \leq f(-5)$, it must be true that $f(-5)=f(5)$.

A sketch of one such graph is below.

Global extrema will either occur at critical points in the interval $(-5,5)$ or at the endpoints $x=5,\,x=-5$.

The global maximum is 45 at $x=5$ and the global minimum is $-19$ at $x=-3$.

Global extrema will occur at critical or singular points in the interval $(-5,5)$ or at the endpoints $x=5,\,x=-5$.

$f'(x)=2x+6$. Since this is defined for all real numbers, there are no singular points. The only time $f'(x)=0$ is when $x=-3$. This is inside the interval $[-5,5]$. So, our points to check are $x=-3,\,x=-5,$ and $x=5$.

The global maximum is 45 at $x=5$ and the global minimum is $-19$ at $x=-3$.

You only need to consider critical points that are in the interval $(-4,0).$

The global maximum over the interval is $61$ at $x=-3$, and the global minimum is $7$ at $x=0$.

Global extrema will occur at the endpoints of the interval, $x=-4$ and $x=0$, or at singular or critical points inside the interval. Since $f(x)$ is a polynomial, it is differentiable everywhere, so there are no singular points. To find the critical points, we set the derivative equal to zero.

The only critical point inside the interval is $x=-3$.

The global maximum over the interval is $61$ at $x=-3$, and the global minimum is $7$ at $x=0$.