College Algebra Chapter 1, Exercises of Algebra

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College Algebra

Chapter 1

Note:

This review is composed of questions similar to those

from the chapter review at the end of chapter 1. This

review is meant to highlight basic concepts from

chapter 1. It does not cover all concepts presented

by your instructor. Refer back to your notes, unit

objectives, handouts, etc. to further prepare for your

exam.

This review is available in alternate formats upon

request.

Find 𝑓 + 𝑔, 𝑓 − 𝑔, 𝑓 ∙ 𝑔,

𝑓

𝑔

for the following pair of functions. State the domain of

each of these functions.

2

𝑓 + 𝑔: Add the functions and combine like terms

2

• 𝑥 + 1 + 3x

2

𝑓 − 𝑔: Subtract the functions and combine like terms

2

• 𝑥 + 1 − 3x

2

𝑓 ∙ 𝑔: Multiply the functions

2

• 𝑥 + 1 )(3x)

3

2

𝑓

𝑔

: Divide the functions

𝑓

𝑔

3 𝑥

2

+𝑥+ 1

3𝑥

(cant be simplified any more)

Domain: Find the domain of f and g first.

Since both are functions that do not contain functions over functions, variables inside roots,

fractional exponents, or negative exponents, the domain for f and g is all real numbers.

This is also the domain for 𝑓 + 𝑔, 𝑓 − 𝑔, 𝑎𝑛𝑑 𝑓 ∙ 𝑔.

To find the domain of

𝑓

𝑔

𝑥 , we need consider the domains of f and g but also that 𝑔 𝑥 ≠ 0.

If x=0, then g(x)=0 which is not allowed. So the domain for

𝑓

𝑔

𝑥 is 𝑥 ≠ 0.

Find the difference quotient of 𝑓 𝑥 = − 2 𝑥

2

+ 𝑥 + 1 ; that is, find

𝑓 𝑥+ℎ −𝑓(𝑥)

ℎ

2

2

2

2

2

2

2

2

2

2

2

2

2

Multiply (x+h)(x+h)

Distribute - 2

Distribute - 1 in front of (− 2 𝑥

2

Factor out an h

Cross out h

Difference quotient is −4𝑥 − 2ℎ + 1

Use the graph of the of function f shown to find:

a) The local min and max

b) The absolute min and max

c) Whether the graph is symmetric to x-axis, the

y-axis, or origin

d) Whether the function is even, odd, or neither

e) Find the intercepts

a) The local minimums are low points: local min is - 1 occurs at x=

local maximums are the high points: local max is 1 occurs at x=- 1

Although the point (4,3) is a high point, there is not a lower point on both sides of it. This is a condition of local

mins and maxes but for absolutes

b) The absolute min is the lowest point of all (unless its an open interval end point): None

The absolute max is the highest point of all (unless its an open interval end point): absolute max is 3 and occurs at

x=

c) Symmetry about the x-axis: No because the graph is not mirrored across the x-axis and every point (x,y), the

point (x,-y) is not on the graph

Symmetry about the y-axis: No because the graph is not mirrored across the y-axis and every point (x,y), the point

(-x,y) is not on the graph

Symmetry about the orgin: No because every point (x,y), the point (-x,-y) is not on the graph. If the graph

continued on after x=4, then we would have had this symmetry.

d) Neither because it is not symmetric about the y-axis and the origin.

e) x-intercepts: - 3,0,3 and y-intercept:

Determine algebraically whether the functions are even, odd, or neither

4 +𝑥

2

1 +𝑥

4

3𝑥

4 +𝑥

6

f(x): Replace x with – x and simplify

2

4

2

4

This is equivalent to 𝑓 𝑥 =

4 +𝑥

2

1 +𝑥

4

so the

equation is even

g(x): Replace x with – x and simplify

6

6

6

This is equivalent to – g(x) so the

equation is odd.

Graph ℎ 𝑥 = (𝑥 − 1 )

2

+ 2 using the techniques of shifting, compressing or stretching, and

reflections. State the domain and based on the graph, find the range.

The parent graph would be 𝑓 𝑥 = 𝑥

2

(shown in blue)

ℎ(𝑥) is obtained by shifting the graph to the right by 1 and up by 2.

There isn’t any compressing, stretching, or reflections.

The domain is all real numbers and the range is [ 2 , ∞)

For the piece-wise function

a) Find the domain c) Find where f(x)= 0

b) Graph the function d) Is f continuous on its domain?

a) The domain (the possible x-values) are found on the right side of the equation.

− 2 ≤ 𝑥 ≤ 1 and 𝑥 > 1

From first inequality we see that - 2 is the smallest number and from the second inequality we

see that ∞ is the largest number.

The domain is − 2 , ∞ or 𝑥 𝑥 ≥ − 2

b) Set each part of the equation equal to zero and solve

Top half:

Since 0 is in the domain of the top portion (− 2 ≤ 𝑥 ≤ 1 ), f(0)=

Bottom half:

x=-1 is not in the domain of the bottom portion (𝑥 > 1 ). So (-1,0) is not an answer to where

f(x)=0.

Continues on next slide…

p is inversely proportional to b. If p= 15 when b is 4, what is p when b is 19. Round to 2

decimal places.

1. Start by writing the equation using k for the constant of proportionality.

𝑘

𝑏

2. Find k using the known b and its corresponding p.

Multiply both sides by 4

3. Using the k=60, put in the b=19 and the k into the equation to solve for p.

A rectangle is inscribed in a circle of radius 7 (see image). Let

P(x,y) be the point in quadrant I that is a vertex of the rectangle

and is on the circle.

a) Express the area A of the rectangle as a function of x.

b) Express the perimeter p of the rectangle as a function of x.

a) Remember that area is length times width.

The width can be expressed as 2x and the length is 2y (see image for details)

Area=2𝑥 ∙ 2𝑦 = 4𝑥𝑦

Since we want the area function to be function of x, we can solve the circle’s

equation for y and substitute this into the y in 4xy.

2

2

2

2

2

Area function is 𝐴 𝑥 = 4𝑥 49 − 𝑥

2

b) Perimeter of a rectangle is twice the length plus twice the width. Again, the width

can be expressed as 2x and the length is 2y.

Perimeter = 2 2𝑥 + 2 2𝑦 = 4𝑥 + 4𝑦

Since we want the perimeter function to be function of x, we can solve the circle’s

equation for y and substitute this in for y. From above, we found 𝑦 = 49 − 𝑥

2

. So

2