Unit 0

Pre-AP Foundations

The algebra and precalc that calculus takes for granted: exact trig values, exponents and logs, rationalizing, identities, composition, graphs and asymptotes.

Topics in this unit

Algebra & Exponent Rules · Exponential Expressions & Equations · Logarithm Rules & Equations · Polynomials & Rational Functions · Functions, Composition & Inverses · Graphs & Parent Functions · Function Transformations · Unit Circle & Exact Values · Trig Identities & Simplification

Algebra & Exponent Rules

Question 1

no calculatorhard

Rationalize and simplify (for

x \neq = 0

\frac{x + 1 - 1}{x}

Show the answer

The answer is D.

Multiply by

\frac{x + 1 + 1}{x + 1 + 1}

: numerator becomes

(x + 1) - 1 = x

, cancel

x

to get

\frac{1}{x + 1 + 1}

Why not A:Multiply by the conjugate $x + 1 + 1$ ; the denominator gets the $+$ sign.
Why not B:After rationalizing, the $x$ cancels and leaves $\frac{1}{x + 1 + 1}$ in the denominator.
Why not C:The $x$ in the numerator cancels with the $x$ from the conjugate product.

Exponential Expressions & Equations

Question 2

no calculatorhard

Evaluate:

2 7^{- 1/3}

Show the answer

The answer is D.

2 7^{1/3} = 3

, and the negative exponent gives the reciprocal:

2 7^{- 1/3} = \frac{1}{3}

Why not A:The negative exponent takes the reciprocal: $2 7^{1/3} = 3$ , so $2 7^{- 1/3} = \frac{1}{3}$ .
Why not B:Negative exponent = reciprocal, not negative: $\frac{1}{3}$ .
Why not C: $2 7^{- 1/3} = \frac{1}{3 27} = \frac{1}{3}$ .

Logarithm Rules & Equations

Question 3

no calculatorhard

Solve for

t

100 e^{0.04 t} = 300

Show the answer

The answer is B.

Divide by 100:

e^{0.04 t} = 3

. Then

0.04 t = ln 3

, so

t = \frac{l n 3}{0.04} = 25 ln 3

Why not A:First divide by 100: $e^{0.04 t} = 3$ , then $t = \frac{l n 3}{0.04} = 25 ln 3$ .
Why not C:Don't forget to divide by $0.04$ : $t = \frac{l n 3}{0.04} = 25 ln 3$ .
Why not D:Divide by 100 first so the right side is $3$ , not $300$ .

Polynomials & Rational Functions

Question 4

no calculatorhard

What are the coordinates of the hole in

f (x) = \frac{x ^{2} - 4}{x - 2}

Show the answer

The answer is C.

\frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2

with a hole at

x = 2

. The

y

-value is

2 + 2 = 4

, so the hole is at

(2, 4)

Why not A:After cancelling, $f = x + 2$ ; at $x = 2$ that gives $y = 4$ , not $0$ .
Why not B: $x = - 2$ is the $x$ -intercept of $x + 2$ , not the hole; the hole is where the factor cancelled, $x = 2$ .
Why not D:Plug $x = 2$ into the simplified $x + 2$ : $2 + 2 = 4$ , so $y = 4$ .

Functions, Composition & Inverses

Question 5

no calculatormedium

The table gives values of

f

and

g

x f (x) g (x) 134251325413542

Find

f (g (2))

Show the answer

The answer is A.

From the table

g (2) = 1

, then

f (1) = 3

Why not B:That's $g (f (2)) = g (5)$ , the wrong order. Do $g (2) = 1$ first, then $f (1) = 3$ .
Why not C: $g (2) = 1$ is the inside step. Now look up $f (1) = 3$ .
Why not D:That's $f (2)$ . Read down the $g$ column first. $g (2) = 1$ , then $f (1) = 3$ .

Question 6

no calculatorhard

Using the table

x f (x) g (x) 134251325413542

find

f (f (4))

Show the answer

The answer is C.

f (4) = 1

, then feed it back:

f (1) = 3

Why not A: $f (4) = 1$ is one application; do it again: $f (1) = 3$ .
Why not B: $4$ is the starting input, not the answer. $f (4) = 1$ , then $f (1) = 3$ .
Why not D:Re-read the $f$ row: $f (4) = 1$ , then $f (1) = 3$ .

Graphs & Parent Functions

Question 7

no calculatorhard

What is the range of the function shown?

Show the answer

The answer is A.

The upper semicircle of radius

2

has lowest value

0

(at the ends) and highest

2

(at the top): range

[0, 2]

Why not B:Only the UPPER half is drawn, so outputs are never negative.
Why not C:The radius is $2$ , so the top of the semicircle is at $y = 2$ , not $4$ .
Why not D:Both $y = 0$ (at the ends) and $y = 2$ (at the top) are attained, so use closed brackets.

Function Transformations

Question 8

no calculatormedium

Which of the following is the graph of

y = - (x + 1)^{2} + 3

Show the answer

The answer is C.

The leading negative makes the parabola open downward;

(x + 1)

shifts the vertex left 1 and

+ 3

shifts it up 3, giving a maximum vertex at

(- 1, 3)

Why not A:This is $y = (x + 1)^{2} + 3$ ; it opens upward because it is missing the negative leading coefficient.
Why not B:This is $y = - (x - 1)^{2} + 3$ , with vertex at $(1, 3)$ ; it is shifted right instead of left.
Why not D:This is $y = - (x + 1)^{2} - 3$ , with vertex at $(- 1, - 3)$ ; it is shifted down instead of up.

Unit Circle & Exact Values

Question 9

no calculatormedium

cos \frac{4 π}{3} = ?

Show the answer

The answer is C.

On the unit circle,

cos \frac{4 π}{3} = - \frac{1}{2}

. Cosine is the

x

-coordinate of the point.

Why not A:This is $sin \frac{4 π}{3}$ . You swapped cosine and sine. Cosine is the $x$ -coordinate.
Why not B:Sign error. Check the quadrant of $\frac{4 π}{3}$ to fix the sign.
Why not D:A special-angle value, but not the one for this angle. Recompute the reference angle and the quadrant sign.

Trig Identities & Simplification

Question 10

no calculatormedium

Using a power-reducing identity,

sin^{2} x

equals:

Show the answer

The answer is A.

Power reduction:

sin^{2} x = \frac{1 - cos 2 x}{2}

(this is the form you integrate in calculus).

Why not B:That's $cos^{2} x$ . For $sin^{2} x$ the sign is MINUS.
Why not C:True, but that's not the power-REDUCED (single-power) form requested.
Why not D:The angle doubles: it's $cos 2 x$ , not $cos x$ .

Question 11

no calculatorhard

Solve on

[0, 2 π)

2 sin^{2} x - sin x - 1 = 0

Show the answer

The answer is D.

Factor as

(2 sin x + 1) (sin x - 1) = 0

sin x = 1 \Rightarrow x = \frac{π}{2}

;

sin x = - \frac{1}{2} \Rightarrow x = \frac{7 π}{6}, \frac{11 π}{6}

Why not A:Factor $(2 sin x + 1) (sin x - 1) = 0$ : also $sin x = - \frac{1}{2}$ gives two more solutions.
Why not B:You dropped the $sin x = 1$ root, $x = \frac{π}{2}$ .
Why not C: $sin x = - \frac{1}{2}$ (not $+ \frac{1}{2}$ ) gives the quadrant III/IV angles $\frac{7 π}{6}, \frac{11 π}{6}$ .

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