Entrepreneurs are concerned with maximizing profits and minimizing losses. Engineers want to minimize shear stress and waste energy production. Chemists wish to maximize the amount of product in an expensive reaction.

Finding **extreme values** of a function is one of the most powerful tools calculus has to offer people working in the real world. We'll showcase only the very basic concepts here to whet your appetite for **derivatives**, a special kind of **limit** explored in great depth starting with Chapter 3 and hinted at in the last unit.

Only basic algebra is needed for this unit.

**vertex**. Find the coordinates of this point without using a graph.

**What goes up must come down.**

Calculus gives us the tools to find extreme values without the use of a graph, as we did in our simple parabola problem. Let's spend some time thinking about how calculus achieves this.

Let's say the parabola of the last problem represents the height of an object thrown straight up into the air. What is the velocity of the object when it reaches the vertex \( (1,5)?\)

**Hint:** You don't have to use any calculations here, just your intuition.

The rate of change of a parabola at a vertex is 0. In fact, the converse is true as well. If a point on a parabola has 0 instantaneous rate of change, it must correspond to the vertex.

Use this to find the smallest possible value of \[ f(x) = 3x^2-2x -1 .\]

What we've done so far doesn't seem very impressive. After all, we have a perfectly good way of finding the vertex of a parabola just by completing the square. But what of a function like \( f(x) = x^3 - 3 x, - \sqrt{3} \leq x \leq \sqrt{3} \) pictured below?

Loading interactive graph...

Sorry, something happened while loading this visualization.
Refresh this page to try again.

From the plot on the left, we see that the graph has both a maximum (highest value) and a minimum (lowest value).

Use the sliders provided to estimate their locations by first zooming to \( z = 0\) and then locating the positions where the line on the right is perfectly horizontal.

In the last problem we used our function microscope to see that the maximum and minimum values of \[ f(x) = x^3-3 x,\quad - \sqrt{3} \leq x \leq \sqrt{3} \] occur at \( \pm 1\) respectively.

Let's see how we can find the same answer *without* the microscope!

Simplify \( \frac{f(b)-f(a)}{b-a} \) as much as possible.

**Hint:** \( b^3 - a^3 = (b-a) \big( b^2 + ab + a^2 \big) \)

Through algebra we found that \( f(x) = x^3-3 x \) implies \( \frac{f(b)-f(a)}{b-a} = b^2 + ab + a^2 -3.\) To find the instantaneous rate of change at \( a,\) take \( b = a.\) Use the result to find the points \(\pm a\) guaranteed to be minima/maxima of \( f(x).\) Enter the value of \( a \) when you are done.

**Remember:** The maximum and minimum values of \( f(x)=x^3-3x\) are achieved when the instantaneous rate of change is 0.

Let's finish this unit with a challenge! The function \( f(x) = \big( x^2-1\big)^2 \) has three points where the instantaneous rate of change vanishes: \( a_1, a_2 , \) and \(a_3.\) Find these values without a graph.

**Hint:** Start by showing that \( \frac{f(b)-f(a)}{b-a} = ( b + a ) \big( b^2 + a^2 -2\big ).\) You might also need
\[ \begin{align} & b^2-a^2 = (b+a)(b-a) \\ & b^4-a^4 = (b^2+a^2)(b+a)(b-a).\end{align}\]

Later chapters will show how we can streamline extreme value computations by **taking the derivative** of a function. We still have much ground to cover, but the essential idea is in place: if we are interested in finding extreme values of a function without a graph, we should look for points where the rate of change vanishes. Rates of change, which are computed with limits, are central to one of the most important calculus applications of all.

In the next unit, we turn to the second major branch of calculus (**integral calculus**) and its connection with limits. We'll ease our way into this vast topic by studying an ancient problem: how can we find the area of a flat object with *curved* edges?

×

Problem Loading...

Note Loading...

Set Loading...