Example 1: Global Extrema

Determine all global extrema of the given function on the interval [-1,3].

$f\left(x\right)=3x^5–20x^3$

Step 1: Find the 1 ^st Derivative, $f^{\prime }\left(x\right)$.

$f\left(x\right)=3x^5–20x^3$

$f\prime \left(x\right)=15x^4–60x^2$

Step 2: Find the critical values of the equation.

$f^{\prime }\left(x\right)=0$ OR $f^{‘}\left(x\right)= \mathit{Does\; Not\; Exist }\left(\mathit{DNE}\right)$

In this situation you have a closed interval, [-4,4], so the endpoints of the interval are locations where $f^{\prime }\left(x\right)= \mathit{Does\; Not\; Exist }\left(\mathit{DNE}\right)$. This gives us two critical values that we must include in our 1 ^st Derivative Test.

Now to find the places where $f^{\prime }\left(x\right)=0$, all that is needed is to take the derivative and set it equal to zero, then solve that equation for x . Most times when you set an equation equal to zero, the algebraic method for solving it is the same. Factor the equation and then set those smaller equations equal to zero, and find their solutions.

Algebra Moves:

i)  Factor a $15x^2$ out of the equation.

ii) Separate it into two math problems with both pieces being set equal to zero.

iii) First, in the left column we will isolate the $x^2$. Start by dividing both sides by $15$,and then take the square root of both sides to solve for x .

iv) Next, in the right column we will again isolate the $x^2$, and then take the square root of both sides to solve for x . When you apply the square root , you get the $\pm$ of that result, not just the positive.

REMEMBER: Any critical value that is not inside the x-interval you are given, [-1,3], gets thrown out, and you do not want to include it in the rest of your process.

$f^{\prime }\left(x\right)= \mathit{Does\; Not\; Exist }\left(\mathit{DNE}\right)$

Closed interval, [-4,4]

${\overline{)x=–4}}^{\mathit{Critical\; Value}} \mathit{and }{\overline{)x=4 }}^{\mathit{Critical\; Value}}$

$f^{\prime }\left(x\right)=0$

$0=15x^4–60x^2$

$0={15x}^2\left(x^2–4\right)$

$0={15x}^2$

$\frac{0}{15}=\frac{{15x}^2}{15}$

$0=x^2$

$\sqrt{0}=\sqrt{x^2}$

$0=x$

${\overline{)0=x}}^{\mathit{Critical\; Value}}$

Keep: ${\overline{)0=x}}^{\mathit{Critical\; Value}}$

$0=\left(x^2–4\right)$

$\begin{array}{c}0\\ \underline{+4}\end{array}=\begin{array}{c}{x}^2–4\\ \underline{+4}\end{array}$

$4=x^2$

$\sqrt{4}=\sqrt{x^2}$

$\pm 2=x$

$–2=\mathit{x }\mathit{and }2=x$

${\overline{)–2=x}}^{\mathit{Critical\; Value}} \mathit{and }{\overline{)2=x\mathrm{ }}}^{\mathit{Critical\; Value}}$

Keep: ${\overline{)2=x\mathrm{ }}}^{\mathit{Critical\; Value}}$

Not in interval

Throw Out: ${\overline{)–2=x}}^{\mathit{Critical\; Value}}$

Step 3: Plug the x-values of your critical values and your endpoints of your closed interval, [ a , b ], back into your original equation, $f\left(x\right)$. This will give you the y-values that go with these x-values .

The nice part about a global extrema problem is we do not need to do the actual 1 ^st Derivative Test (number line game). We just need to find the y-valuesthat go with these x-values .

The x-values we will use for this process will be:

1)      Critical Value: x = 0

2)      Critical Value: x = 2

3)      End Point: x = -1

4)      End Point: x = 3

$f\left(x\right)=3x^5–20x^3$

1)       $f\left(0\right)=3{\left(0\right)}^5–20{\left(0\right)}^3=0$

2)       $f\left(2\right)=3{\left(2\right)}^5–20{\left(2\right)}^3=–64$

3)       $f\left(–1\right)=3{(–1)}^5–20{(–1)}^3=17$

4)       $f\left(3\right)=3{\left(3\right)}^5–20{\left(3\right)}^3=189$

Step 4: Draw conclusions.

The largest y-value , is the winner of the Absolute/Global Max award.

The smallest y-value , is the winner of the Absolute/Global Min award.

1)       $f\left(0\right)=3{\left(0\right)}^5–20{\left(0\right)}^3=0$

2)       ${\overline{)f\left(2\right)=3{\left(2\right)}^5–20{\left(2\right)}^3=–64}}^{\mathit{Global }\mathit{Min}}$

3)       $f\left(–1\right)=3{(–1)}^5–20{(–1)}^3=17$

4)       ${\overline{)f\left(3\right)=3{\left(3\right)}^5–20{\left(3\right)}^3=189}}^{\mathit{Global }\mathit{Max}}$

Final Result:

On the interval [-1,3], the function $f\left(x\right)=3x^5–20x^3$ has a global min of y = -64 at x = 2 .

On the interval [-1,3], the function $f\left(x\right)=3x^5–20x^3$ has a global max of y = 189 at x = 3 .