## Relative vs Absolute Extrema

Although a relative extrema may seem to be very similar to an absolute extrema, they are actually quite different. The term “relative” means compared to numbers nearby…so a relative extrema is either a bump or a dip on the function.

The term “absolute” means the most extreme on the entire function. An absolute extrema is the very highest or lowest point on the function. This may occur at a bump or a dip. They may also occur at the ends of the function if it is defined on a closed interval.

The MathFAQ below illustrates how to find these points on a function.

Goto the MathFAQ >>

## It’s All Relative…or It’s Not

Extrema are the high and low points on a function’s graph.

It can be difficult to distinguish between the bumps and dips on a graph, the relative (or local) extrema, and the very highest and lowest points on a graph, the absolute extrema.

The Math-FAQ belows show how to find each type of extrema using derivatives.

Goto the Math-FAQ >>

## How Do You Estimate a Derivative at a Point vs Compute a Derivative at a Point from the Limit?

We are now in the real meat of calculus….rates and derivatives. Essentially, rates are simply slopes. Depending on the field you are working in or the representation of the data or function, it might have a different terminology. Keeping all of these straight is the biggest challenge in the class. Let’s try to put what we know into a table.

Common Description Graphical Interpretation Mathematical Terminology
Average Rate of Change of f(x) from x=a to x=b Slope between (a,f(a)) and (b,f(b))  Slope of the Secant Line from x=a to x=b
Instantaneous Rate of Change of f(x) at x=a  Slope between (a,f(a)) and (a+h,f(a+h)) where h is very small  Slope of the Tangent Line on f(x) at x=a

If it feels like we use slope to calculate everything…you are right. The only difference is where the two points are located. If they are located fairly far apart, then we are calculating an average rate of change. If they are located an infinitesimal amount apart, then we are calculating the instantaneous rate of change.

But here is where it gets hazy…if the representation of the function does not allow us to pick the points infinitely close together, we may approximate the instantaneous rate of change with an average rate of change between two points that are as close as the representation allows. I think this point is bothering many of you. It occurs most often when we try to find the instantaneous rate from a data table. In that case, the points are where they are at and we can pick pairs that are closer and closer together. We pick them as close as we can near the place we want the instantaneous rate of change and say we are estimating the instantaneous rate of change.

If we are given a graph, we can draw a secant line between the points to calculate the average rate of change. The only estimating done in this case is estimating where the points are on the graph. Our eyes are only so good in reading off the values. For the instantaneous rate of change, we draw the tangent line where we want the rate and eyeball its slope. Again, since we are making educated guesses about the slopes, the numbers are estimates based on our ability to read the slope.

If we are given the function’s formula, we can calculate the average rate or the instantaneous rate exactly. For the average rate of change of f(x) over x = a to x = b, we calculate ${{f(b) - f(a)} \over {b - a}}$.

For the instantaneous rate of change, we use the function’s formula to calculate the limit $\mathop {lim }\limits_{h \to 0} {{f(a + h) - f(a)} \over h}$.

The estimating comes from the fact that we may not be able to find two points infinitesimally close or the fact that we cannot calculate the slope perfectly from how the function is given to us.

Another cautionary note…different disciplines call the derivative by different names. Mathematicians call it the derivative of course, but in physics it might be called the instantaneous rate. In finance and economics, it will be called the marginal function. And in other fields you’ll here other terms.

## Do You Have Even More Substitution Examples?

The in-person class has spent some time at the board doing substitution problems. I took pictures so that you would have more examples handy. Several of these examples are similar to ones you all have had trouble with on the homework.

There is a lot of good work here…any mistakes? Which solution is easiest to read…and perhaps the best?

Example 1    $\displaystyle \int{\left( x+1 \right){{\left( {{x}^{2}}+2x \right)}^{2}}dx}$

Example 2     $\displaystyle \int{4x{{e}^{{{x}^{2}}+9}}dx}$

Example 3     $\displaystyle \int{6x{{e}^{2{{x}^{2}}+1}}dx}$

Example 4      $\displaystyle \int{5x{{e}^{10{{x}^{2}}+4}}dx}$

Example 5     $\displaystyle \int{\frac{\sqrt{2+\ln \left( x \right)}}{x}dx}$

Example 6     $\displaystyle \int{\left( {{x}^{3}}+2x \right){{\left( {{x}^{4}}+4{{x}^{2}}+7 \right)}^{8}}dx}$

Example 7      $\displaystyle \int{\left( 6-6z \right){{e}^{2z-{{z}^{2}}}}dz}$

Example 8     $\displaystyle \int{\left( x+1 \right){{\left( {{x}^{2}}+2x \right)}^{3}}dx}$

Example 9     $\displaystyle \int{\left( 3{{x}^{2}}+4 \right){{\left( 2{{x}^{3}}+8x \right)}^{19}}dx}$

## How Do You Find Quantities Related to Revenue and Cost?

In this problem, you were given revenue and cost functions like $\displaystyle R(x)=-{{x}^{2}}+8x$ and $\displaystyle C(x)=2x+5$. In the parts of the problem you were asked to a) identify the correct graph, b) find the break-even quantity, c) the maximum revenue, and d) the maximum revenue. In class yesterday, the students worked in groups to solve this problem. Here is the work they put on the board.

Parts a and b

Parts c and d

Look pretty good but on part d, they should write the profit at x = 4 as P(4). The key in thee part is to use $\displaystyle x=-\frac{b}{2a}$ on the quadratic revenue and profit functions to find the quantities at which the maximums are achieved. Since the parts asked you to find the maximum revenue and profit, you need to put the quantities into the functions.