Colinearity

1 Co linearity
Answer: No

Are the fol lowing points colinear?
{(3, 1), (4, 4), (5, 8)}

Take the slope of two points , say (3, 1) and (4, 4) we compute the slope:

Now create the equation joining those two points, use Point-Slope Formula using either of
the two points we used of your choice (here I use (3, 1)):

Now look at the point we did not pick, (5, 8). If these three points are colinear, I should be
able to plug in x = 5, and get y = 8. However.
y = 3(5) - 2 ⇒ y = 13

which is not what we are looking for, therefore these 3 points are not colinear.

2 Solving Inequalities

2.1  x² - 9  ≤ 0
Answer: [-3,3]

2.2  x² - 2x - 8 ≥ 0
Answer:
x² - 2x - 8 ≥ 0 ⇒ (x - 4)(x + 2) ≥0

Since we want positive values , we need to see when both factors are the same sign . Break
inequality into 2 parts : x - 4 ≥ 0 and x + 2 ≥ 0
Solving for x for both of these gives us x ≥ 4 and x ≥ -2.
Note that for any given x x ≥ 4 and x ≥ -2 must be satisfied, hence we choose x ≥ 4.
That is not all however. x² - 2x - 8 ≥ 0 is also positive when both x - 4 and x + 2 are
negative.
So solve x - 4 ≤ 0 and x + 2 ≤ 0
We get x ≤ 4 and x ≤ -2.
As with before for any given x x ≤ 4 and x ≤ -2 must be satisfied, hence we chose x ≤ -2.
Since our solutions are: x ≥ 4 and x ≤ -2, our interval notation would be:

2.3 6|x - 3| < 18
Answer: (0,6)
Divide both sides by 6, then rewrite as: -3 < x - 3 < 3 and
solve to get 0 < x < 6.

2.4 2|x - 4| > 10
Answer:
Divide both sides by 2 and since we are dealing with >, we need split into 2 expressions :
x - 4 > 5 and -(x - 4) > 5
Solving for x gives us two solutions: x > 9 and x < -1
Which in interval notation is:

2.5 x² + 3 < 0
Answer:
As you do it out, you nd that x² < -3 is impossible.
In this course you can't take a negative square root

2.6 |e^x + 9| < 0
Answer:
As you do it out, you nd that e^x < -9 is impossible.
Can't take the ln of a negative number

3 Equations of Parallel or Perpendicular Lines

3.1
Find an equation of a line Parallel to y = 3(x - 2) + 1 through given point (0, 3)
If we want a parallel line, then we know that m = 3 and we have the point (0, 3)
Just use point slope from here y - 3 = 3(x - 0)
and we get y = 3x + 3

3.2
Find an equation of a line Perpendicular to y = 2x + 1 through given point (3, 1)
If we want a perpendicular line, then we know that and we have the point (3, 1)
Just use point slope from here, and we get

4 Functions (Domain, Range, Inverse, Composition)

Given f(x) = x² + 2 and
Domain of f(x) is all R or   since we can plug anything in.
Range of f(x) is [2,) which can be noted by observing the graph . Notice that the vertex
of the graph (absolute minimum in this case) is at (0,2). So the range can never be less than 2.

Domain of g(x) is [-2,) since we cannot have a negative square root.
Range of g(x) is [0,)

Assuming common domain, [-2,] you will notice these are inverses of eachother.

Thus,

All that needed to be done here was perform the indicated operation with the 2 functions

To show they are inverses of eachother we need to show f ◦ g = g ◦ f = x
For f ◦ g we are taking g(x) and putting it into f(x) which is also denoted by f(g(x)). So:

For g ◦ f we are taking f(x) and putting it into g(x) which is also denoted by g(f(x)). So:


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