# Graph the system of equations

Q:  Graph the system of linear equations, determine the number of solutions, and find the solution:

y = 4x + 2 and y =- 2x – 3

A: Okay, for the sake of talking about the lines I will name them:

Line 1:  y = 4x + 2

Line 2:  y =- 2x – 3

A “solution” to a system of linear equations is the point where the lines intersect.  So, we need to graph both of these lines on the same grid and see if and where they intersect.

Step 1:  Graph Line 1 and Line 2 on the same grid:

Tricks for graphing line 1:  Since line 1 is in the form y = mx + b, we can use the method of finding the slope and the y-intercept to graph the line.

The slope of line 1 is 4 and the y-intercept of line 1 is 2.

On your x-y graph, go up 2 on the y-axis and put a point [this is the y-intercept, or the “b” value]

Now, starting from the y-intercept point 2:  go up 4 and right 1 (this is the slope 4/1).  Create another point.  Connect the dots to make a line.  You should have: Now, we need the same strategy to graph line 2 on the same grid.

Line 2:  y =- 2x – 3

The y-intercept is -3 and the slope is -2.  Plot the y-intercept of -3 on the y-axis.  From that point:  go down 2, right 1 to create another point.  Connect the dots to make a line.  Your graph should now look like: So, the question now is:  How many solutions are there and what is the solution.  Recall, the solution is (x, y) point of intersection.
Clearly, there is 1 solution.  The lines do cross at 1 point.

What is that point?  Well, you have to estimate that point from the graph.  This is one problem with solving a system by graphing — it can be inaccurate.  Depending on the “neatness” of your graph, your answer will vary.

To me, the answer appears to be around the point (-.8, -1.3)

*To get the accurate solution, we solve these types of problems algebraically.

# Finding the limit example

Q: Find the limit (as x approaches 3) of (x3 – 27) / (3-x)

A:  The first thing to do when finding a basic limit is try plugging in the number in question (3).

So, plug in 3 to get:

(33 – 27) / (3-3) = 0/0 <– if you get 0/0 or infinity/infinity that means there is more work to be done.  However, if you had just got a number like 4 or something, that would’ve been your answer!

OK, we got 0/0 so that means more work.  More work could mean many things (apply different rules, factor and cancel, simplify, etc).  In this case, it appears we can factor, so we try that:

(x3 – 27) / (3-x) = (x – 3)(x2 + 3x + 9)/ (3 – x)

Now, here comes some tricky insight.  I notice that the (x – 3) on top is very similar to the (3 – x) on the bottom.  I am going to factor a “-1” out of the (x – 3) that is on top.

Notice:  -1(3 – x) = (x – 3)

So, the numerator becomes:

-1(x – 3)(x2 + 3x + 9)/ (3 – x)

Now, the (x – 3) term cancels from the top and bottom to leave:

-1(x2 + 3x + 9)

So, we are trying to find the limit (as x approaches 3) of -1(x2 + 3x + 9).  We have “removed the hole” — the factor (3 – x) was a “removable hole” that was causing calculation problems.  In the simplified version, we can plug in the value 3 to now calculate where that hole was occuring:

Lim (x –> 3) of -1(x2 + 3x + 9) = -1(32 + 3(3) + 9) = -1(9 + 9 + 9) = -27

# Negative Exponent Examples and Basics

Q:  What is a negative exponent?  What does a negative exponent do?

A:  A negative exponent is just notation that can be used to represent a reciprocal.  Let me show you some examples:

x – 1 = 1/x1 or 1/x

x-2 = 1/x2

4 – 1 = 1/41 or 1/4

4 – 2 = 1/42 = 1/16

Those are the basics.  Let’s look at a few more complicated examples of how negative exponents affect an expression:

3x – 4 = 3/x4

4x6y – 3 = 4x6/y3

OR… If the negative exponent is affecting something in a denominator, it will move it to the numerator like so:

1/x – 5 = 1*x5 = x5

6/y – 7 = 6y7

These are just a few basic examples to show you how negative exponents work. Of course things can get more difficult.

# Quotient Rule Example

Q:  Find dy/dx of

y = x / sqrt(x2 + 1)

A:  To find dy/dx (the derivative), we will need to use the quotient rule since we have a function over a function.  See? We are in the form:

y = f / g where f and g are two different functions of x.

In this form, the quotient rule tells us if:

y = f / g
then

y ‘ = (g * f ‘ – f * g ‘ ) / g2

So, we know:

f = x

g = sqrt(x2 + 1)

f ‘ = 1   <— basic derivative

g ‘ = (this takes more work…. First, rewrite “g”):

g = sqrt(x2 + 1) = (x2 + 1)1/2

Now, use a power rule and a chain rule to find g ‘ like so:

g ‘ = 1/2 (x2 + 1) – 1/2 (2x)

[that’s the derivative of the outside * the derivative of the inside]

Clean up g ‘:

g ‘ = x (x2 + 1) – 1/2

So now we have all of the players:  f, f ‘ , g, g ‘:

f = x

g = (x2 + 1)1/2

f ‘ = 1

g ‘ = x (x2 + 1) – 1/2

Now, we said:  y ‘ = (g * f ‘ – f * g ‘ ) / g2, so plug in the pieces then simplify like so:

y ‘ = [ (x2 + 1)1/2(1) – (x)(x (x2 + 1) – 1/2 ) ] / [(x2 + 1)1/2]2

CLEAN UP

y ‘ = [ (x2 + 1)1/2 – x2(x2 + 1) – 1/2  ] / (x2 + 1)

Multiply top and bottom by (x2 + 1)1/2

[ (x2 + 1)1/2(x2 + 1)1/2 (x2 + 1)1/2x2(x2 + 1) – 1/2  ] / (x2 + 1)1/2(x2 + 1) =

[ (x2 + 1)- x2 ] / (x2 + 1)3/2

1 / (x2 + 1)3/2

So, if:

y = x / sqrt(x2 + 1)

dy/dx = 1 / (x2 + 1)3/2

Q:  Solve for x:  (x + 3)(x – 4) < 0

A:

Step 1:  Find the zeroes

Since the quadratic is already factored, this isn’t too tricky.  If it wasn’t factored, you’d have to factor first! (always make sure there is a 0 on one side of the equation / inequality before proceeding).

OK, so what are the zeroes?

x + 3 = 0 or x – 4 = 0

The zeroes are x = -3 or x = 4

So, draw a number line and plot the zeroes on the number line: Now, you have to test each “interval” that is separated by the “zeroes”.  There are three intervals to test.

Interval 1:  The numbers to the left of -3  –> in interval notation this is (-infinity, -3)

Interval 2:  The numbers between -3 and 4  –> in interval notation this is (-3, 4)

Interval 3:  The numbers to the right of 4  –> in interval notation this is (4, infinity)

Step 2:  Test each interval

Pick any number on interval 1 and test it into the original inequality.  I’ll pick -5:

(x + 3)(x – 4) < 0

(-5 + 3)(-5 – 4) < 0

(-2)(-9) < 0

18 < 0      <—  this is false, so numbers on this interval [interval 1] are not part of the solution.

Pick any number on interval 2 and test it into the original inequality.  I’ll pick 0:

(x + 3)(x – 4) < 0

(0 + 3)(0 – 4) < 0

(3)(-4) < 0

-12 < 0      <—  this is true, so numbers on this interval [interval 2] are a part of the solution.

Pick any number on interval 3 and test it into the original inequality.  I’ll pick 5:

(x + 3)(x – 4) < 0

(5 + 3)(5 – 4) < 0

(8)(1) < 0

8 < 0      <—  this is false, so numbers on this interval [interval 3] are not part of the solution.

SO:  The only interval that “worked” was interval 2.  Therefore, the solution is all number between -3 and 4.

In interval notation, we write that like: (-3, 4)

In inequality notation, we write that like: -3 < x < 4

# Minimum / Maximum, Concavity, Etc!

Q:  Consider the function:  f(x) = (1/3) x3 – x2 – 15x + 1

(a) critical points

(b) the local minimum / maximums [relative extreme]

(c) points of inflection

(d) concavity

A:  OK.  Get out your pen and paper and be writing with me as I calculate and explain.  This is the best way for you to learn.

(a)  The critical points are where the derivative equals 0 or is undefined.  So, take the derivative to start:

f(x) = (1/3) x3 – x2 – 15x + 1

f ‘ (x) = x2 – 2x- 15  <– you should get this!

Now, we need to find where the derivative = 0 (or is undefined — since the derivative is a quadratic, it is never undefined, so we don’t have to worry about that part).

0 = x2 – 2x- 15

Solve by factoring:

0 = (x – 5)(x + 3)

So, the critical points are when x = 5, -3.

(b)  To determine if the critical points are local maximum or minimums, we need to do the second derivative test.

Recap on second derivative test:  Plug a critical point into the second derivative.

*If the end result is positive, this means the shape is concave up (smiley face) which means we have a minimum.

*If the end result is negative, this means the shape is concave down (frowny face) which means we have a maximum.

So, take the second derivative:

f(x) = (1/3) x3 – x2 – 15x + 1  [original function]

f ‘ (x) = x2 – 2x- 15  [1st derivative]

f ” (x) = 2x – 2  [2nd derivative]

Now, we have critical points of x = 5 and x = -3.  Deal with them one at a time in the second derivative:

f ” (5) = 2(5) – 2 = 10 – 2 = 8 <– this is positive, which means when x = 5 we have a local minimum!

f ” (-3) = 2(-3) – 2 = -6 – 2 = -8 <– this is negative, which means when x = -3 we have a local maximum!

If you wanted to find the (x, y) point of the actual max and min, you would plug the x-values into the original equations to find the corresponding y values:

f(5) = (1/3) (5)3 – (5)2 – 15(5) + 1 = -172/3

So, f(x) has a local minimum at the point (5, -172/3)

f(-3) = (1/3) (-3)3 – (-3)2 – 15(-3) + 1 = 28

So, f(x) has a local maximum at the point (-3, 28)

(c)  The points of inflection are where the concavity of the function changes from concave up to concave down (and vice versa).  Our POI candidates can be found by setting the second derivative = 0 and finding the x candidates:

f ” (x) = 2x – 2

0 = 2x – 2 and solve for x:

x = 1

So, x = 1 is a candidate for a point of inflection (POI) — it is not a guaranteed POI though.

We need to interval test into the second derivative to see what is happening (generally, you would need to test in between every x value.  We only have 1 x value, so we just need to interval test on either side of it):

Plot your x values on a number line.  In our case, the only value is x = 1.  Pick a number on the left side of 1.  Let’s pick an easy number, 0:

Plug 0 into the second derivative:

f ” (0) = 2(0) – 2 = -2 <– since this is negative:  this interval, to the left of x = 1, is concave down

Now, pick a number to the right of x = 1.  Let’s pick 2:

f ” (2) = 2(2) – 2 = 2 <– since this is positive:  this interval, to the right of x = 1, is concave up

Therefore, when x = 1 we do have a point of inflection (since we changed from concave down to concave up).

If you want the (x, y) point, plug the x value into the original equation to find y:

f(1) = (1/3) (1)3 – (1)2 – 15(1) + 1 = -44/3

There is a point of inflection at (1, -44/3).

(d)  Now to analyze concavity, which we have basically already done in part (c)!

To the left of x = 1, we determined the graph to be concave down.

To the right of x = 1, we determined the graph to be concave up.

Concave down on the interval (-infinity, 1)

Concave up on the interval (1, infinity)
We have done it.  That was a lot to do.  Enjoy.

# Rolle’s Theorem: Concept and Example

Q:  Can you explain Rolle’s Theorem?

A:  Rolle’s Theorem is a specific application of the Mean Value Theorem.

What you need to apply Rolle’s Theorem:  a function that is differentiable everywhere on the interval you are looking at.  Basically, you need a smooth function (rolling hills, no points, breaks, sharp corners, holes, etc…)

Rolle’s says that if the function attains the same height at two different points (and meets the “smooth function” criteria) then there must be a point in the middle somewhere where the derivative of the function is zero.

Example:

Consider the function (I just made it up for example sake):
f(x) = x2 – 8x

Now, I know:

*This is a smooth / differentiable function

*And, f(2) = -12 and f(6) = -12  [just plug in 2 and then plug in 6 to verify… 2 and 6 are numbers I picked for the sake of the example]

So, Rolle’s Thm tells me that since f(2) and f(6) both = -12 and the function is differentiable…. There is an x value somewhere between 2 and 6 where the derivative of the function is 0.  Essentially, there is a “turn around point”  In order to go from the point (2, -12) to the point (6, -12), you must eventually “turn around” <— which gives you a derivative of 0.

That is all Rolle’s tells you.  It doesn’t tell you what the point is (and it doesn’t tell you if there is just 1 point where this happens or if there are more).  Rolle’s just tells you there is at least 1 point in that interval where the derivative will be 0.  You have to do the work to find that point (or points).

Let’s do it for fun:

f(x) = x2 – 8x

Take the derivative:

f ‘ (x) = 2x – 8

Set = 0 to find the turnaround point:

0 = 2x – 8

Solve for x (I won’t show those steps):

x = 4.

So, the derivative is 0 when x = 4.  And of course, x = 4 is definitely on the interval from 2 to 6 as Rolle’s predicted.

***Keep in mind, this is a very basic example.  It turned out the the derivative only had 1 zero anyway.  The derivative could have had MANY zeroes…. Rolle’s just would have guaranteed us that one of them (or more) was on our interval!

# Linearization: Concept and Example

Q:  Find the linearization of f(x) = ex at x = 0

A:  First, some concept:  Linearization is the act of finding a “linear function” that can approximate the given function on or around a given point.  In this problem, we want to find a line that models the shape of ex when you are around the point x = 0.

Step 1:  Find an (x, y) point on the function in question.

The function is f(x) = ex .  The x part of the point is 0.  Plug that in to find y:

f(0) = e0 = 1

So, the point is (0, 1).  Hold this point.  We will need it for later.

Essentially, we want to find a line that follows that patterns of f(x) = ex and goes through the point (0, 1).

So, to model the pattern, we need to slope of f(x) = ex at the given point.

Step 2:  Find the derivative (slope) of the function at the given point:

f(x) = ex

Find the derivative:

f ‘ (x) = ex

Now, find the slope at (0, 1):

f ‘ (0) = e0 = 1

So, the slope is 1.  m = 1

Step 3:  Find a line that has the same slope as the function that goes through the given point:

We need a line with slope of 1 that goes through the point (0, 1).

y = mx + b

We know m = 1, so:

y = x + b

Plug in the (x, y) point to find “b”

1 = 0 + b

1 = b

So, the equation of the line is y = x + 1

In summation, to approximate values of f(x) = ex around where x = 0 you can use the line y = x + 1

Q:  Solve  (a/2)2 – a > 0

Solving Algebraically:

Step 1:  Simplify

We can simplify the equation by squaring the fraction to get:

a2/4 – a > 0

OR in different form:

(1/4) a2 – a > 0

Now, to get rid of the fraction and clean things up, I am going to multiply everything by 4 (since the fraction is 1/4 — you don’t have to do this, just a “cleaning up” step”):

4*(1/4) a2 – 4*a > 4*0

a2 – 4a > 0

Normally, when you solve inequalities, you isolate the variable by moving things around to the left / right side.  When it is a quadratic, you don’t want to do that.  You want all of the numbers and variables on one side and zero on the other side.  We have that, so we are good to go.

Step 2:  Factor

Now, factor the left side:

a2 – 4a > 0

a (a – 4) > 0

Step 3:  Identify the zeroes

As we are used to doing with quadratics, we need to find what values you plug in to make “zeroes”

So, take each factor and set it equal to zero like so:

a = 0 and a – 4 = 0

Solve to get:

a = 0 and a = 4

The zeroes are:  0 and 4.

So, this parabola (quadratic) crosses the x-axis at 0 and 4.  Now, we want to find where the quadratic is greater than 0.

With analysis, we know the quadratic is concave up in shape (a smiley face).  On your paper, draw an x-y graph with a parabola that crosses the x-axis and 0 and 4 and is concave up.  You should get: So, where is the parabola > 0 (above the x-axis?)

When a < 0 and when a > 4

# Graphing Lines

Q:  Graph x + 2y = 4

A:  We are going to graph this by making an x-y table of values.  A typical (though not necessary) thing to do is solve for y first — this just means “get y by itself”

x + 2y = 4

Subtract x from both sides (to try to get y alone)

2y = 4 – x

Now, it’s standard for to list the x first, so we may write this as:

2y = -x + 4

To get y by itself, divide everything by 2:

y = -x/2 + 4/2

This simplifies to:

y = -1/2 x + 2

So, to make a table of values, pick some numbers for x.  It is standard to pick “0, 1, 2” or something similar.  So:

x  |  0  |  1  |  2  |

y  |      |      |      |

Plug in each value for x to find y:

When x = 0, y = -1/2 (0) + 2 = 0 + 2 = 2

When x = 1,  y = -1/2 (1) + 2 = -1/2 + 2 = 1.5

When x = 2, y = -1/2 (2) + 2 = -1 + 2 = 1

Then the table is complete:

x  |  0  |  1    |  2  |

y  |  2  | 1.5  |  1  |

These are just points you need to plot on a graph:  (0, 2)  (1, 1.5)  and (2,1)

Plot the points then connect the dots to make a line!