Question

Global extrema on an interval: The first-derivative test can be used to show that the function $f\left(x\right)=x^2+3x$has a local minimum at $x=-\frac{3}{2}.$Is this a global minimum of the function? Is there a global maximum? What are the global extrema (if any) if we consider the function restricted to the interval [−3, 3]?

Short answer

$$\begin{gathered} x=-\frac{3}{2}isthelocalminimumofthegivenfunction. \\ Thegivenfunctiondoesnothaveanyglobalmaximum. \\ x=-\frac{3}{2}istheglobalminimumintheinterval[-3,3]forthefunctiongiven. \end{gathered}$$

Step-by-step solution

Step 1. Given Information.

Given the function: $f\left(x\right)=x^2+3x$.

Step 2. First-derivative test.

Firstly we need to find the critical points:

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ Since,f\left(x\right)=x^2+3x, \\ 2x+3=0 \\ x=-\frac{3}{2}. \\ Itisthecriticalpoint. \end{gathered}$$

$$\begin{gathered} Now,f\text{'}\text{'}\left(x\right)=2>0, \\ thismeansx=-\frac{3}{2}isalocalminima. \end{gathered}$$

There is not any global maximum.

If we consider the interval [-3,3] then we get,

$$\begin{gathered} f(-3)={\left(-3\right)}^2+3(-3)=9-9=0 \\ f\left(3\right)=3^2+3\left(3\right)=9+9=18 \\ f\left(-\frac{3}{2}\right)={\left(-\frac{3}{2}\right)}^2+3\left(-\frac{3}{2}\right)=\frac{9}{4}-\frac{9}{2}=-\frac{9}{4}. \\ Nowfromabovewecanseethatf\left(-\frac{3}{2}\right)isminimum. \\ So,x=-\frac{3}{2}istheglobalminimumintheinterval[-3,3]. \end{gathered}$$