Q. 1 - Calculus

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]?

Step-by-Step Solution

Answer

$$\begin{gathered} x=-\frac{3}{2} is the local minimum of the given function. \\ The given function does not have any global maximum. \\ x=-\frac{3}{2}is the global minimum in the interval [-3,3] for the function given. \end{gathered}$$

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:

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

$$\begin{gathered} Now, f\text{'}\text{'}\left(x\right) = 2>0, \\ this means x=-\frac{3}{2}is a local minima. \end{gathered}$$

There is not any global maximum.

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

$$\begin{gathered} f(-3) = {\left(-3\right)}^2+3(-3) = 9-9=0 \\ f\left(3\right) = 3^2+3\left(3\right) = 9+9=18 \\ f\left(-\frac{3}{2}\right) = {\left(-\frac{3}{2}\right)}^2+3\left(-\frac{3}{2}\right) = \frac{9}{4}-\frac{9}{2}=-\frac{9}{4}. \\ Now from above we can see that f\left(-\frac{3}{2}\right) is minimum. \\ So, x=-\frac{3}{2}is the global minimum in the interval [-3,3]. \end{gathered}$$