Differential Calculus Pretest Solutions

## Solution to Question 1

1. It is easier to determine the slope, y-intercept, and x-intercept of a linear line if it is in the form:

(1)
\begin{equation} y = mx +b \end{equation}

Where m refers to the slope, and b refers to the y-intercept. This is known as the slope-intercept form of a linear line. The line given was in the point-slope form of a linear line, so we should convert it to assist us in finding what we need by expanding the -4 through to the (x - 2), and by isolating y:

(2)
\begin{equation} y - 5 = -4(x - 2) \end{equation}
(3)
\begin{equation} y - 5 = -4x + 8 \end{equation}
(4)
\begin{equation} y = -4x + 8 + 5 \end{equation}
(5)
\begin{equation} y = -4x + 13 \end{equation}

We can now easily acknowledge the slope is equal to -4, and the y-intercept is equation to 13. To find the x-intercept, we must set y = 0 and isolate x.

(6)
\begin{equation} 0 = -4x + 13 \end{equation}
(7)
\begin{equation} -13 = -4x \end{equation}
(8)
\begin{align} \frac{-13}{-4} = x, \end{align}
(9)
\begin{align} \frac{13}{4} = x, \end{align}

So the answer to question 1 follows:

(10)
\begin{align} x-intercept: (\frac{13}{4}, 0), y-intercept: (0, 13), m = -4 \end{align}

## Solution to Question 2

Remember that to find an equation of a linear line, we need a set of points and a slope. We are given a set of points, but we will need the slope. We can acquire that with the given equation. Let's first convert it to y = mx + b form to extract the slope.

(11)
\begin{equation} 0 = -2y + 8x -6 \end{equation}
(12)
\begin{equation} 2y = 8x -6 \end{equation}
(13)
\begin{equation} y = 4x -3 \end{equation}

We can now see that the slope is equal to 4.

Note that the question asks for an equation of a line that is perpendicular to the one given. Hence, remember that perpendicular lines have //negative reciprocal slopes, meaning the slope will be written as thus generally given

(14)
\begin{equation} m = a \end{equation}

Then:

(15)
\begin{align} m_{\perp} = \frac{-1}{a} \end{align}

Since the slope of the original line is 4:

(16)
\begin{equation} m = 4 \end{equation}

Then we know that any line perpendicular to the original line has a slope of:

(17)
\begin{align} m_{\perp} = \frac{-1}{4} \end{align}

So we have that the slope of the perpendicular line is -1/4, and the point is must pass through is (2,6). Now we can insert this information into the point-slope form of a linear line:

(18)
\begin{equation} y - y_{0} = m(x - x_{0}) \end{equation}

Thus giving us the following equation when we let x0 = 2, y0 = 6, and m=-1/4:

(19)
\begin{align} y - 6 = \frac{-1}{4} (x - 2) \end{align}

Similarly to question 1, we can now expand the equation out to get y = mx + b form.

(20)
\begin{align} y - 6 = \frac{-1}{4}x - \frac{-1}{4}2 \end{align}
(21)
\begin{align} y - 6 = \frac{-1}{4}x + \frac{1}{2} \end{align}
(22)
\begin{align} y = \frac{-1}{4}x + \frac{1}{2} + 6 \end{align}
(23)
\begin{align} y = \frac{-1}{4}x + \frac{1}{2} + \frac{12}{2} \end{align}
(24)
\begin{align} y = \frac{-1}{4}x + \frac{13}{2} \end{align}

Alternatively, you could also not use point-intercept form at all, but rather use slope-intercept form to solve for the y-intercept and plug the values in directly to the equation.

For example, given that we have a set of points (2,6) and the slope of the perpendicular line being -1/4, we can isolate y = mx + b to solve for b:

(25)
\begin{equation} y = mx + b \end{equation}
(26)
\begin{align} 6 = \frac{-1}{4} 2 + b \end{align}
(27)
\begin{align} 6 = \frac{-1}{2} + b \end{align}
(28)
\begin{align} 6 + \frac{1}{2}= b \end{align}
(29)
\begin{align} \frac{13}{2}= b \end{align}

Now that we know b = 13/2, and m = -1/4, we can put those directly into y = mx + b to receive the exact same answer.

(30)
\begin{align} y = \frac{-1}{4}x + \frac{13}{2} \end{align}

## Solution to Question 3

The quadratic formula is necessary for this question because it cannot be factored by breaking up the term containing x. Thus, there are no factors of ac which their sum equals to b.

Now, we shall assign values for a, b, and c. To do so, the quadratic equation must have all x-terms on one side. In this case, it is already done for us. We shall assign a to be coefficient of x^2. We shall assign b to be the coefficient of x. And we shall assign c to be the constant term, or the number by itself. Thus we get:

(31)
\begin{equation} a = 3, b = -5, c = 7 \end{equation}

Now we will apply the quadratic formula:

(32)
\begin{align} x = -b \pm \frac{\sqrt{b^2 - 4ac}}{2a} \end{align}

To get:

(33)
\begin{align} x = -(-5) \pm \frac{\sqrt{(-5)^2 - 4(3)(7)}}{2(3)} \end{align}

When simplified, we get:

(34)
\begin{align} x = 5 \pm \frac{\sqrt{25 - 84}}{6} \end{align}
(35)
\begin{align} x = 5 \pm \frac{\sqrt{-59}}{6} \end{align}

The problem is there is a negative value under the square-root/radical. Because of this, there are zero x-intercepts.

Note that while this question does not ask for the number of x-intercepts, an easy way to find the number is through the determinant. The determinant is known as:

(36)
\begin{equation} b^2 - 4ac \end{equation}

Or simply, the value under the radical. The following holds true:

(37)
\begin{align} if \quad b^2 - 4ac < 0 \end{align}

Then there are no x-intercepts as the value under a radical cannot be less than 0 or it will result in an imaginary number.

(38)
\begin{align} if \quad b^2 - 4ac = 0 \end{align}

Then there is exactly one x-intercept as x = 0. Note that some texts may state there being "two equal roots", but this means the same thing as 0 = 0.

(39)
\begin{align} if \quad b^2 - 4ac < 0 \end{align}

Then there are exactly two x-intercepts. We already concluded our original quadratic function has no x-intercepts however.

Moving on to y-intercepts, we will let x = 0. This is usually easy as the y-intercept is common to be equal to c.

(40)
\begin{equation} y = 3x^2 -5x + 7 \end{equation}
(41)
\begin{equation} y = 3(0)^2 -5(0) + 7 \end{equation}
(42)
\begin{equation} y = 7 \end{equation}

Thus our y-intercept is as (0, 7)

We then need to find the vertex of the equation. This is rather simply given that the vertex is equal to:

(43)
\begin{align} v = \frac{-b}{2a} \end{align}

A proof of the following will be referenced below as supplemental learning. Nevertheless, we can determine the vertex, v, to be:

(44)
\begin{align} v = \frac{-(-5)}{2(3)} \end{align}
(45)
\begin{align} v = \frac{5}{6} \end{align}

So our vertex is located at x = 5/6

To determine if our quadratic opens up or down, we need to look at the sign associated with a. Because a = 3, and thus a is positive, then the quadratic equation opens up. Below is a graph of the equation: ## Solution to Question 4

To determine the roots of the equation, let's first factor out a 4:

(46)
\begin{equation} 4x^2 - 36 \end{equation}
(47)
\begin{equation} 4(x^2 - 9) \end{equation}

We should now recognize that x-squared subtract 9 can be easily factored because it is a difference of squares. Differences of squares have the following form of factorization:

(48)
\begin{align} (x - y)^2 = (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) \end{align}

Thus, we can factor further to get:

(49)
\begin{equation} 4(x + 3)(x - 3) \end{equation}

Therefore, we have solutions at x = 3, and x = -3.

## Solution to Question 5

To simplify, let's try factoring the numerator of:

(50)
\begin{align} \frac{x^2 - 6x + 9}{x - 3} \end{align}

When factoring the numerator, we get:

(51)
\begin{align} \frac{(x-3)(x-3)}{x - 3} \end{align}

We can verify it similarly as we factored above. We know a = 1, b = -6, and c = 9. So are there factors of a * c that are equal to be b?

Well a * c = 9. The factor sets of 9 are: (1,9), (-1,-9), (3,3), and (-3,-3). But (-3) + (-3) = b = -6! So we can break apart the numerator to get:

(52)
\begin{equation} x^2 - 3x -3x + 9 \end{equation}

Then factor each two terms separately:

(53)
\begin{equation} x(x - 3) -3(x - 3) \end{equation}

And then we can factor out an (x - 3) to get:

(54)
\begin{equation} (x - 3)(x - 3) \end{equation}

For the numerator. Now back to the simplified function:

(55)
\begin{align} \frac{(x-3)(x-3)}{x - 3} \end{align}

We can now remove an (x - 3) term from the numerator and the denominator to get:

(56)
\begin{equation} x - 3 \end{equation}

As the simplified form. To figure out the equations of vertical asymptotes, we must look at the original function and determine where x does not exist.

(57)
\begin{align} \frac{(x-3)(x-3)}{x - 3} \end{align}

It is clear that x =/= 3, because if x = 3, then the denominator will result in 0. Nothing can be divided by zero for our purposes. So the equation of the vertical asymptote is x = 3.

## Solution to Question 6

To do this question, let's first draw a diagram for easy visualization and determine what the rectangle should look like. The blue rectangle is formed from the four given lines. To find the area of the rectangle, we will need a length and a width. The length of one of the sides is clearly 7 - 3 = 4. The length of the other side is 2 - (-6) = 8. Thus, 4*8 = 32. You may count the number of miniature 1x1 squares to verify, but it is not practical for larger questions.

## Solution to Question 7

This question is a basic recall question. You should know the unit circle by heart. You will only ever be expected to know the basic unit circle angles without the use of a calculator.

(58)
\begin{align} sin(60^\circ) = \frac{\sqrt{3}}{2} \end{align}
(59)
\begin{align} tan(\frac{5 \pi}{6}) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{align}
(60)
\begin{align} csc(30^\circ) = \frac{1}{tan(30^\circ)} = \frac{1}{ \frac{1}{\sqrt{3}} } = \sqrt{3} \end{align}

## Solution to Question 8

There are two ways to do this question. One way is with the unit circle and with the CAST rule. We know that sinx is positive in quadrants 1 and quadrants 2. We also know that cosx is positive in quadrants 1 and quadrants 4. Hence, sinx = cosx has only one unique solution which is in quadrant 1. This only occurs when x = 45 degrees according to the unit circle.

(61)
\begin{align} sin(45^\circ) = cos(45^\circ) \end{align}

Another way of doing this is algebraically, though some answers may be missed.

(62)
\begin{equation} sin(x) = cos(x) \end{equation}
(63)
\begin{align} \frac{sin(x)}{cos(x)} = 1 \end{align}
(64)
\begin{align} tan(x) = 1, x = 45^\circ \end{align}

Remember, there is only one unique solution, however, there are infinite solutions because the unit circle can revolve around the origin infinite times in both a clockwise and counterclockwise direction. For that reason, x = 45deg + 360deg is also a solution, and x = 45 -360 deg is also a solution. It is thus, necessary to write a general solution such that:

(65)
\begin{align} x = 45^\circ \pm n360^\circ, n \in I \end{align}

The notation may look funky, but all the above expression states is that x can equal 45 degrees plus or minus any integer value multiplied by 360 degrees. Note that integers are …-2, -1, 0, 1, 2…

## Solution to Question 9

Remember the unit circle identity such that:

(66)
\begin{equation} cos^2(x) + sin^2(x) = 1 \end{equation}

It can be arranged such that

(67)
\begin{equation} cos^2(x) = 1 - sin^2(x) \end{equation}

But nothing that this is present within the identity, so we can substitute it in for simplification to get:

(68)
\begin{align} tan(x) \frac{cos^2(x)}{(sin(x))^2} = cot{x} \end{align}

Now, remember the following rule:

(69)
\begin{equation} sin^2(x) = (sin(x))^2 \end{equation}

Not that the following rule is not true but is commonly mistaken as true:

(70)
\begin{equation} sin^2(x) =/= sin(x^2) \end{equation}

We can now adjust the equation to get:

(71)
\begin{align} tan(x) \frac{cos^2(x)}{sin^2(x)} = cot{x} \end{align}

We know that cosx/sinx = cotx, thus we can apply this identity:

(72)
\begin{equation} tan(x) cot^2(x) = cot{x} \end{equation}

But remember that cotx = 1/tanx, thus:

(73)
\begin{align} \frac{tan(x)}{tan(x)tan(x)} = cot{x} \end{align}

We can also cancel out a pair of tan(x) in the numerator and the denominator because tanx/tanx = 1 to get:

(74)
\begin{align} \frac{tan(x)}{tan(x)tan(x)} = cot{x} \end{align}

And convert the left side to cot(x) and we've proven the identity.

(75)
\begin{equation} cot(x) = cot{x} \end{equation}

## Solution to Question 10

Remember that the domain refers to all possibly values for which a function can exist. A lot of these are just a look and state sort of questions, however, the use of limits in the future will be necessary to solve harder and more complex function domains.

For the equation:

(76)
\begin{equation} y = /sqrt{(x + 3)} \end{equation}

We know that any value in the square root cannot be a negative number (for now). Thus, the minimum value that x can be is -3 inclusive. Hence, x must be greater or equal to -3 for this function, and the domain can be written in interval notation as:

(77)
\begin{align} D: [-3, \infty) \end{align}

Or it may also be written in terms of an inequality such that

(78)
\begin{align} -3 \leq x < \infty \end{align}

Note that the square-bracket corresponds to the less-than-or-equal OR greater-than-or-equal symbols, while the circle bracket corresponds to the less-than or greater-than symbols. Also note that the square bracket implies inclusion, while the round bracket does not. There is NEVER a square bracket on positive infinity or negative infinity.

The domain for the next may be a little trickier. We know that for ln(x), that:

(79)
\begin{equation} 0 < x \end{equation}

Note that x cannot be 0, but must be greater than zero. Hence, let's set the inside portion of the natural logarithm to zero.

(80)
\begin{equation} 2x - 3 = 0 \end{equation}
(81)
\begin{equation} 2x = 3 \end{equation}
(82)
\begin{equation} x = 1.5 \end{equation}

Thus, we know that when x = 1.5, then the inside of the natural logarithm is zero. Hence the domain does not include 1.5 or any values less than 1.5 and can be written in interval notation in the following.

(83)
\begin{align} D: (1.5, \infty) \end{align}

Once again, pay attention to the type of brackets! Both are round brackets this time.

We will not bother to show the inequality notation because it is a bit redundant.

Lastly the final domain question is a little bit trickier. We can look at the domains of the individual terms though. Note that 2x has a domain from negative infinity to positive infinity. The domain of e to the power of x also has a domain from negative infinity to positive infinity. Therefore:

(84)
\begin{align} D: (-\infty, \infty) \end{align}

## Solution to Question 11

This question requires logarithm rules. When we factor 128 we get:

(85)
\begin{equation} y = log_{2}(16*4*2) \end{equation}

With log laws, we can separate this such that:

(86)
\begin{equation} y = log_{2}(16*4*2) = log_{2}(16) + log_{2}(4) + log_{2}(2) \end{equation}

These summation of the logs on the right side should be easy now to compute.

(87)
\begin{equation} y = log_{2}(16) + log_{2}(4) + log_{2}(2) = 4 + 2 + 1 \end{equation}

Thus

(88)
\begin{equation} y = 7 \end{equation}

## Solution to Question 12

First, let's use the log law for coefficients of logarithms to remove the 2 in front of the natural logarithm in the exponent. We thus get:

(89)
\begin{equation} x = e^{ln(5^2)} \end{equation}

And further expanded we get:

(90)
\begin{equation} x = e^{ln(25)} \end{equation}

Now we should be familiar that e^x and ln(x) are inverses of each other, hence, x = 25.