Finding Extrema on Closed Intervals

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We show a procedure to find global extrema on closed intervals.

From the previous section, we know that if a function is continuous on a closed interval then it achieves a global maximum and global minimum on that interval. If we further know that this function is differentiable on this interval, then we have the following observation:

Suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ . Then $f$ has a global maximum at a critical point or an endpoint and a global minimum at a critical point or endpoint

We can use this observation in the following way:

Find all critical points for $f$ that lie in the given interval
Evaluate $f$ at each critical point and at the endpoints
The global maximum is the largest value such produced and the global minimum is the smallest such value produced

It is important to note that we are evaluating the original function $f$ at the critical points and not $f'$ . Moreover, only the critical points that are contained in the given interval are to be checked.

Find the global maximums and minimums for the function $f(x)=x^3-12x$ over the interval $[-3,5]$

We begin by finding the critical points for $f$ $\ddx f(x)=\answer [given]{3x^2-12}.$ This is defined everywhere and is zero at $x=\pm 2$ . So the critical points are when $x=\pm 2$ . Now we evaluate $f$ at all critical points as well as the endpoints of this interval. $\begin{align*} f(-2) &= \answer[given]{16} \\ f(2) &= \answer[given]{-16} \\ f(-3) &= \answer[given]{9} \\ f(5) &= \answer[given]{65} \end{align*}$

Since the largest value that occurs is $65$ , so $65$ is the global maximum for $f$ over $[-3,5]$ and occurs at the endpoint $x=5$ . The smallest value of $f$ that occurs is $-16$ , so $-16$ is the global minimum for $f$ over $[-3,5]$ and occurs at the critical point $x=2$ .

Find the global maximums and minimums for the function $f(x)=x^3-12x$ over the interval $[-1,3]$

We begin by finding the critical points for $f$ $\ddx f(x)=\answer [given]{3x^2-12}.$ This is defined everywhere and is zero at $x=\pm 2$ . So the critical points are when $x=\pm 2$ . Notice that the critical point $x=-2$ is not in the interval $[-1,3]$ , therefore we do not check at $x=-2$ Now we evaluate $f$ at all critical points in $[-1,3]$ as well as the endpoints of this interval. $\begin{align*} f(-1) &= \answer[given]{11} \\ f(2) &= \answer[given]{-16} \\ f(3) &= \answer[given]{-9} \end{align*}$

Since the largest value that occurs is $11$ , so $11$ is the global maximum for $f$ over $[-1,3]$ and occurs at the endpoint $x=-1$ . The smallest value of $f$ that occurs is $-16$ , so $-16$ is the global minimum for $f$ over $[-1,3]$ and occurs at the critical point $x=2$ .