Composition of Functions: Inverse Functions and Composition (page 6 of 6). Sections: Composing functions that are sets of point, Composing functions at points, Composing functions with other functions, Word problems using composition, Inverse functions and composition. So, 2 2 would be typed 2^2. X 2 would be typed x^2. (x+5) 2 would be typed (x+5)^2. You can put a fraction in an exponent. X 2/3 should be typed like x^(2/3). With more complicated fractions you have to use parenthesis. For example if you typed x^2+1/x-5, you might think this means 'the quantity 'x-squared plus 1' over the quantity 'x minus 5.

1. Soulver 3 0 4 X 2
2. Soulver 3 0 4 X 200

Graphing Linear Inequalities

Learning Objective(s)

·Determine whether an ordered pair is a solution to an inequality.

·Graph an inequality in two variables.

Linear inequalities can be graphed on a coordinate plane. The solutions for a linear inequality are in a region of the coordinate plane. A boundary line, which is the related linear equation, serves as the boundary for the region. You can use a visual representation to figure out what values make the inequality true—and also which ones make it false. Let’s have a look at inequalities by returning to the coordinate plane.

Linear Inequalities as Regions

Linear inequalities are different than linear equations, although you can apply what you know about equations to help you understand inequalities. Inequalities and equations are both math statements that compare two values. Equations use the symbol =; inequalities will be represented by the symbols <, ≤, >, and ≥.

One way to visualize two-variable inequalities is to plot them on a coordinate plane. Here is what the inequality x > y looks like. The solution is a region, which is shaded.

There are a few things to notice here. First, look at the dashed red boundary line: this is the graph of the related linear equation x = y. Next, look at the light red region that is to the right of the line. This region (excluding the line x = y) represents the entire set of solutions for the inequality x > y. Remember how all points on a line are solutions to the linear equation of the line? Well, all points in a region are solutions to the linear inequality representing that region.

Let’s think about it for a moment—if x > y, then a graph of x > y will show all ordered pairs (x, y) for which the x-coordinate is greater than the y-coordinate.

The graph below shows the region x > y as well as some ordered pairs on the coordinate plane. Look at each ordered pair. Is the x-coordinate greater than the y-coordinate? Does the ordered pair sit inside or outside of the shaded region?

The ordered pairs (4, 0) and (0, −3) lie inside the shaded region. In these ordered pairs, the x-coordinate is larger than the y-coordinate. These ordered pairs are in the solution set of the equation x > y.

The ordered pairs (−3, 3) and (2, 3) are outside of the shaded area. In these ordered pairs, the x-coordinate is smaller than the y-coordinate, so they are not included in the set of solutions for the inequality.

The ordered pair (−2, −2) is on the boundary line. It is not a solution as −2 is not greater than −2. However, had the inequality been x ≥ y (read as “x is greater than or equal to y'), then (−2, −2) would have been included (and the line would have been represented by a solid line, not a dashed line).

Let’s take a look at one more example: the inequality 3x + 2y ≤ 6. The graph below shows the region of values that makes this inequality true (shaded red), the boundary line 3x + 2y = 6, as well as a handful of ordered pairs. The boundary line is solid this time, because points on the boundary line 3x + 2y = 6 will make the inequality 3x + 2y ≤ 6 true.

As you did with the previous example, you can substitute the x- and y-values in each of the (x, y) ordered pairs into the inequality to find solutions. While you may have been able to do this in your head for the inequality x > y, sometimes making a table of values makes sense for more complicated inequalities.

If substituting (x, y) into the inequality yields a true statement, then the ordered pair is a solution to the inequality, and the point will be plotted within the shaded region or the point will be part of a solid boundary line. A false statement means that the ordered pair is not a solution, and the point will graph outside the shaded region, or the point will be part of a dotted boundary line.

Graphing Inequalities

So how do you get from the algebraic form of an inequality, like y > 3x + 1, to a graph of that inequality? Plotting inequalities is fairly straightforward if you follow a couple steps.

Let’s graph the inequality x + 4y ≤ 4.

To graph the boundary line, find at least two values that lie on the line x + 4y = 4. You can use the x- and y- intercepts for this equation by substituting 0 in for x first and finding the value of y; then substitute 0 in for y and find x.

Plot the points (0, 1) and (4, 0), and draw a line through these two points for the boundary line. The line is solid because ≤ means “less than or equal to,” so all ordered pairs along the line are included in the solution set.

The next step is to find the region that contains the solutions. Is it above or below the boundary line? To identify the region where the inequality holds true, you can test a couple of ordered pairs, one on each side of the boundary line.

If you substitute (−1, 3) into x + 4y ≤ 4:

This is a false statement, since 11 is not less than or equal to 4.

On the other hand, if you substitute (2, 0) into x + 4y ≤ 4:

This is true! The region that includes (2, 0) should be shaded, as this is the region of solutions.

And there you have it—the graph of the set of solutions for x + 4y ≤ 4.

A quick note about the problem above. Notice that you can use the points (0, −3) and (2, 1) to graph the boundary line, but that these points are not included in the region of solutions, since the region does not include the boundary line!

Summary

When inequalities are graphed on a coordinate plane, the solutions are located in a region of the coordinate plane, which is represented as a shaded area on the plane. The boundary line for the inequality is drawn as a solid line if the points on the line itself do satisfy the inequality, as in the cases of ≤ and ≥. It is drawn as a dashed line if the points on the line do not satisfy the inequality, as in the cases of < and >. You can tell which region to shade by testing some points in the inequality. Using a coordinate plane is especially helpful for visualizing the region of solutions for inequalities with two variables.

Just as 2 by 2 matrices can represent transformations of the plane, 3 by 3 matrices can represent transformations of 3D space. The picture is more complicated, but as in the 2 by 2 case, our best insights come from finding the matrix's eigenvectors: that is, those vectors whose direction the transformation leaves unchanged.

If non-zero {bf e} is an eigenvector of the 3 by 3 matrix A, then

for some scalar lambda . This scalar is called an eigenvalue of A.

This may be rewritten

and in turn as

As in the 2 by 2 case, the matrix A-lambda ,I must be singular. Once again, then, we ask: which are the values of lambda for which A-lambda ,I is singular? That is, the values that satisfy the characteristic equation

Consider the example

The characteristic equation is

Expanding the determinant,

Expanding the brackets and simplifying:

or, equivalently

By trial and error, we find that

and it follows from the Factor Theorem that left(lambda -3right) is a factor. Indeed,

Soulver 3 0 4 X 2

and

meaning that the eigenvalues are 3, -5 and 6.

We now go on to solve

for each eigenvalue lambda . Now, every such system will have infinitely many solutions, because if {bf e} is an eigenvector, so is any multiple of {bf e}. So our strategy will be to try to find the eigenvector with X=1, and then if necessary scale up. (If there is no such eigenvector, we know that X must in fact be zero, and we instead look for the eigenvector with Y=1, and so on.)

Eigenvector corresponding to eigenvalue 3

In the case lambda =3, we have

Setting X=1 gives, as our first two equations,

Subtracting the first from the second:

and thus Y=-frac{3}{2}.

Substituting back into the second equation,

giving Z=-frac{1}{2}.

Checking in the third equation,

which works. This gives us the eigenvector

For convenience, we can scale up by a factor of 2, to get

Eigenvector corresponding to eigenvalue -5

Soulver 3 0 4 X 200

In the case lambda =-5, we have

Setting X=1 gives, as our first two equations,

Subtracting the first from the second:

and thus Y=frac{1}{2}.

Substituting back into the second equation,

giving Z=-frac{1}{2}.

Checking in the third equation,

which works. This gives us the eigenvector

Once again, we can scale up by a factor of 2, to get

Eigenvector corresponding to eigenvalue 6

In the case lambda =6, we have

Setting X=1 gives, as our first two equations,

Subtracting the first from the second:

and thus Y=6.

Substituting back into the second equation,

giving Z=16.

Checking in the third equation,

which works. This gives us the eigenvector

General considerations

In general, the eigenvalues of a real 3 by 3 matrix can be

(i) three distinct real numbers, as here;

(ii) three real numbers with repetitions;

(iii) one real number and two conjugate non-real numbers.

The geometric interpretation of the transformation depends on which of the above is true: the first will involve stretches in the three eigenvector directions, the third will involve a rotation and a stretch along its axis and the second will usually involve one of several types of 3D shear. We leave these details for the moment, however.

Another important use of eigenvalues and eigenvectors is diagonalisation, and it is to this that we now turn.