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Subsection 3.5.2 Finding Global Maxima and Minima

We now have a technique for finding local maxima and minima — just look at endpoints of the interval of interest and for values of \(x\) for which either \(f'(x)=0\) or \(f'(x)\) does not exist. What about finding global maxima and minima? We'll start by stating explicitly that, under appropriate hypotheses, global maxima and minima are guaranteed to exist.

So let's again consider the question

Suppose that the maximum (or minimum) value of \(f(x)\text{,}\) for \(a\le x\le b\text{,}\) is \(f(c)\text{.}\) What does that tell us about \(c\text{?}\)

If \(c\) obeys \(a \lt c \lt b\) (note the strict inequalities), then \(f\) has a local maximum (or minimum) at \(x=c\) and Theorem 3.5.4 tells us that either \(f'(c)=0\) or \(f'(c)\) does not exist. The only other place that a maximum or minimum can occur are at the ends of the interval. We can summarise this as:

This theorem provides the basis for a method to find the maximum and minimum values of \(f(x)\) for \(a\le x\le b\text{:}\)

Let's now demonstrate how to use this strategy. The function in this first example is not too simple — but it is a good example of a function that contains both a singular point and a critical point.

Find the largest and smallest values of the function \(f(x)=2x^{5/3}+3x^{2/3}\) for \(-1\le x\le 1\text{.}\)

Solution We will apply the method in Corollary 3.5.13. It is perhaps easiest to find the values at the endpoints of the intervals and then move on to the values at any critical or singular points.

  • Before we get into things, notice that we can rewrite the function by factoring it:
    \begin{align*} f(x) &= 2x^{5/3}+3x^{2/3} = x^{2/3} \cdot \left(2x + 3\right) \end{align*}
  • Let's compute the function at the endpoints of the interval:
    \begin{align*} f(1) &= 2 +3 = 5\\ f(-1) &= 2 \cdot(-1)^{5/3} + 3\cdot (-1)^{2/3} =-2 + 3 = 1 \end{align*}
  • To compute the function at the critical and singular points we first need to find the derivative:
    \begin{align*} f'(x) &= 2 \cdot \frac{5}{3} x^{2/3} + 3 \cdot \frac{2}{3} x^{-1/3}\\ &= \frac{10}{3} x^{2/3} + 2 x^{-1/3}\\ &= \frac{10 x + 6}{3 x^{1/3}} \end{align*}
  • Notice that the numerator and denominator are defined for all \(x\text{.}\) The only place the derivative is undefined is when the denominator is zero. Hence the only singular point is at \(x=0\text{.}\) The corresponding function value is
    \begin{align*} f(0) &= 0 \end{align*}
  • To find the critical points we need to solve \(f'(x) = 0\text{:}\)
    \begin{align*} 0 &= \frac{10 x + 6}{3 x^{1/3}} \end{align*}
    Hence we must have \(10x=-6\) or \(x=-3/5\text{.}\) The corresponding function value is
    \begin{align*} f(x) &= x^{2/3} \cdot \left(2x + 3\right) & \text{recall this from above, then}\\ f(-3/5) &= (-3/5)^{2/3} \cdot\left(2 \cdot \frac{-3}{5} + 3 \right)\\ &= \left(\frac{9}{25}\right)^{1/3} \cdot \frac{-6 + 15}{5}\\ &= \left(\frac{9}{25}\right)^{1/3} \cdot \frac{9}{5} \approx 1.28 \end{align*}
    Note that if we do not want to approximate the root (if, for example, we do not have a calculator handy), then we can also write
    \begin{align*} f(-3/5) &= \left(\frac{9}{25}\right)^{1/3} \cdot \frac{9}{5}\\ &= \left(\frac{9}{25}\right)^{1/3} \cdot \frac{9}{25} \cdot 5\\ &= 5 \cdot \left( \frac{9}{25} \right)^{4/3} \end{align*}
    Since \(0 \lt 9/25 \lt 1\text{,}\) we know that \(0 \lt \left( \frac{9}{25} \right)^{4/3} \lt 1\text{,}\) and hence
    \begin{gather*} 0 \lt f(-3/5) = 5 \cdot \left( \frac{9}{25} \right)^{4/3} \lt 5. \end{gather*}

  • We summarise our work in this table

    \(c\) \(-\frac{3}{5}\) \(0\) \(-1\) \(1\)
    type critical point singular point endpoint endpoint
    \(f(c)\) \(\frac{9}{5}\root{3}\of{\frac{9}{25}}\approx 1.28\) \(0\) \(1\) \(5\)
  • The largest value of \(f\) in the table is \(5\) and the smallest value of \(f\) in the table is \(0\text{.}\)
  • Thus on the interval \(-1\leq x \leq 1\) the global maximum of \(f\) is \(5\text{,}\) and is taken at \(x=1\text{,}\) while the global minimum value of \(f(x)\) is \(0\text{,}\) and is taken at \(x=0\text{.}\)

  • For completeness we also sketch the graph of this function on the same interval.

    Later (in Section 3.6) we will see how to construct such a sketch without using a calculator or computer.

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