Lab 7: Max/Min’s

0. Objective:

This lab will illustrate how Mathematica can be used to find absolute and local maximum and minimum values of a function on a closed interval.  If you have not already done so, read section of the text on Absolute Maximums and Minimums.

1. Examples:

Consider the following function on the interval [-5, 4].

f[x_]=x^3-3*x^2-9*x+7

The critical values of this function can be found by the following Mathematica command

cv=Solve[f'[x]==0,x]

The function values at these critical values are

f[x]/.cv

So, f(-1) = 12 and f(3) = -20.  We also need to know the function values at the endpoints of the interval:

f[{-5,4}]

Thus, the absolute minimum is at -5 and has a value of -148 and the absolute maximum value is at -1 and has a value of 12.  We can look at a graph to verify this:

Plot[f[x],{x,-5,4}]

As another example, let’s look at the following function:

g[x_]=((((x-1))^3)/(((x+1))^2))

Using Mathematica to calculate the derivative gives us:

Simplify[g'[x]]

It is easy to see from this that the function has critical values (in the domain) of -5 and 1.  To see if these c.v. are local extreme, look at graphs near the c.v.:

Plot[g[x],{x,-6,-4}]

Plot[g[x],{x,0,2}]

From this, it looks like the function has a local maximum at -5, but that the c.v. of 1 is neither a max nor a min.

Name:
PIN:

Task:

Consider the function h(x)=(((x^2)(x-5))^(2/3)) on the interval [-2.1, 5.2].

(1) Use Mathematica to help you sketch a graph of this function on the given interval.  Note that you may have to be creative on how you enter the "two-thirds" power to see the entire graph.

(2) Find all critical values and classify the critical values as being locations for either a local maximum or local minimum or neither.

(3) Find the absolute extreme values of the function h on the interval [-2.1, 5.2] accurate to at least 3 decimal digits.