Precalculus Quiz 1 study guide

A Relation

a set of ordered pairs like: (x,y). The set of first elements in the ordered pairs is the domain, while the set of second elements is the range.

State the domain and range of the following relation: {(5,2), (30,8), (15,3), (17,6), (14,9)

Domain: {5,30,15,17,14} → {5,14,15,17,30}

Range: {2,8,3,6,9} →{2,3,6,8,9}

The domain of a relation is all non zero positive integers less than 6. The range y of the relation is 3 less x, where x is a member of the domain. Write the relation as a table of values and as an equation. Then graph the relation.

Table:

x	y
1 2 3 4 5	2 1 0 -1 -2

Equation:

y=3-x

Graph:

A Function

a relation in which each element of the domain is paired with exactly one element in the range.

NO REPEATED DOMAIN

State the domain and range of each relation. Then state whether the relation is a function. No two pairs have the same first element.

a.) {(-2,1), (3,-1), (2,0)} →

b.) {(3,-1), (3,-2), (9,1)} →

a.) Domain: {-2, 3, 2} Range: {1, -1, 0}

A function because the Domain isn’t repeated.

b.) Domain: {3, 3, 9} Range: {-1, -2, 1}

Not a function because there is a repeated Domain

Evaluate each function for the given value:

a.) f(-4) if f(x) = 3x²-7x²-2x

b.) g(9) if g(x) = |6x-77|

c.) g(x+2) if g(x) = 2x-5

d.) f(c-5) if f(x) = x²-7x-4

a.) -296

b.) 23

c.) 2x-1

d.) c²-17c+56

State the domain of each function:

**In any fraction the denominator can’t be zero*

a.) f(x) = \frac{3}{x-1}

b.) g(x) = \frac{3-x}{5+x}

c.) f(x) = \sqrt{1x} \sqrt{\left(1\right)x}

a.) x - 1 ≠ 0

Domain: x ≠ 1

b.) 5 + x ≠ 0

Domain: x ≠ -5

c.) x ≠ any negative number. Because you can’t square root ( \sqrt{\left(1\right)x} ) a negative #

Domain: x ≥ 0

State the domain and range of each relation. The state whether the relation is a function or not. Explain.

1. {(1,2),(2,4),(3,6),(0,0)}

2. {(6,2),(3,4),(6,6),(3,0)}

1. Domain: {1,2,3,0}

   Range: {2,4,6,0}

   It is a Function because the Domain has no repeating numbers.

2. Domain: {6,3,6,3}

   Range: {2,4,6,0}

   It is NOT a Function because the Domain is repeating the number 6 and 3 twice.

State the domain and range. and if it is a function or not.

x	y
-3 0 3 6	4 0 -4 8

Domain: {-3,0,3,6}

Range: {4,0,-4,8}

This is a Function, in every input there is only one output.

Evaluate each function for the given value:

13. f(-3) if f(x) = 4x³+x²-5x

14. g(m+1) if g(x) = 2x²-4x+2

13. f(-3) if f(x) = 4x³+x²-5x

    4(-3)³ + (-3)²-5(-3)

    → f(-3)= -84

14. g(m+1) if g(x) = 2x²-4x+2

    2(m+1)²-4(m+1)+2

    2(m²+2m+1)-4m+4+2

    2m²+4m+2-4m-4+2

    2m²+2-2

    → g(m1) = 2m²

Given the table, evaluate the following.

1. f(1) = x

2. f(-2) = x

3. f(2) = x

4. f(x) = -7

5. f(x) = -5

x	y
0 1 2 -1 -2	-5 -3 -1 -7 -9

1. f(1) =-3

2. f(-2) = -9

3. f(2) = -1

4. f(-1) = -7

5. f(0) = -5

Given f(x) = 3x²-4 and g(x) = 4x+5, find each function.

a.) \left(f+g\right)\left(x\right)

b.) \left(f-g\right)\left(x\right)

c.) \left(f\cdot g\right)\left(x\right)

d.) \left(\frac{f}{g}\right)\left(x\right)

e.) \left(f+g\right)\left(3\right)

a.) \left(f+g\right)\left(x\right)

f(x)+g(x)

3x²-4+4x+5

= 3x²+1+4x

b.) \left(f-g\right)\left(x\right)

f(x)-f(x)

3x²-4-4x+5

= 3x²-4x+1

c.) \left(f\cdot g\right)\left(x\right)

f(x) \cdot g(x)

(3x²-4)(4x+5)

= 12x³+15x²-16x-20

d.) \left(\frac{f}{g}\right)\left(x\right)

=\frac{3x^2-4}{4x+5}

e.) \left(f+g\right)\left(3\right)

f(3) + g(3)

3(3)² -4 + 4(3)+5

27-4+12+5

23+17

= 40

Given f(x) = x²+7 and g(x) = x-3, find (f ∘ g)(x) and (g ∘ f)(x)

a) g(x) = x-3

f(g)= f(x-3)

f(x-3) = (x-3)²+7

f(x-3) = (x-3)(x-3)+7

f(x-3) = x²-3x-3x+9+7

f(x-3) = x²-6x+16

b) f(x)= x²+7

g(f) = g(x²+7)

g(x²+7) = x²+7-3

g(x²+7) = x²+4

Perform the indicated operation.

1) f(x) = 4x+5

Find f(f(-1))

2) f(x) = x+5

g(x) = -x²-5x

Find f(g(-4))

3) g(n) = n+1

f(n) = n²-n

Find g(f(8))

4) g(x) = 2x

Find g(g(-3))

5) g(n) = 4n

f(n) = 4n+2

Find g(f(n))

6) g(x) = 3x+3

f(x) = x³-3x²

Find g(f(x))

1) 9

2) a) 4

b) 9

3) a) 56

b) 57

4) a) -6

b) -12

5) 16n+8

Domain: \left(-\infty,\infty\right)

6) 3x³-9x²+3

Domain: \left(-\infty,\infty\right)

Adding and Subtracting Functions

Let f(x) = 4x+7 and g(x) = 12+x

What are f+g and f-g?

a) 4x+7+12+x

= 5x +19

b) 4x+7-(12+x)

= 5x-5

Multiplying and Diving Functions

Let f(x) = x² -9 and g(x) = x+3

What are f\cdot g and \frac{f}{g}

a) (x²-9) (x+3)

= x³+3x²-9x-27

b) \frac{x^2-9}{x+3}

Evaluate each composite value

1. If f(x) = 3x-5 and g(x) = x², find (f ∘ g)(3) and (g ∘ f)(3)

2. If f(x) = -9x-9 and g(x) = \sqrt{x-9} , find (f ∘ g)(10) and (f ∘ f)(-2)

3. If f(x) = -4+2 and g(x) = \sqrt{x-8} , find (f ∘ g)(12) and (f ∘ f)(5)

1. a.) (i) 9 (ii) 22

   b.) (i) 4 (ii) 16

2. a.) (i) \sqrt1 (ii) -18

   b.) (i) 9 (ii) -90

3. a.) (i) 2 (ii) -18

   b.) (i) -6 (ii) 74

Given f(x) = -9x-9+3 and g(x) = , findx^4(f ∘ g)(x)

f(g(x)) = g(x) = x^4

f(x) = -9x^4 + 3

Given f(x) = 2x-5 and g(x) = x+2, find (f ∘ g)(x) and (g ∘ f)(x)

a.) (f ∘ g)(x) =

f(g(x)) = x+2

2(x+2)-5

2x+4-5

f(x) = 2x-1

b.) (g∘ f)(x) =

g(f(x)) = 2x-5+2

g(x) = 2x- 3

Given g(x) = x²+6x and f(x) = 3x+1, find f(g(x)) and g(f(x))

a.) f(g(x)) =

x² + 6x

f(x) = 3(x²+6x)+1

f(x) = 3x² + 18x + 1

b.) g(f(x)) =

f(x) = 3x+1

g(x) = (3x+1)² + 6(3x+1)

g(x) = 9x²+6x+1+18x+6

g(x) = 9x² +24x + 7

Let g(x) = x-3 and h(x) = x²+6. What is (h ∘ g)(1)?

10

Domain

The set of all x - values that makes the function defined

Polynomial Function -

The DOMAIN of a Polynomial Function is always the set of Real Numbers. In a polynomial function the exponent must be positive.

Polynomial Function: Domain of f(x) = 3x-5

all real numbers or \left(-\infty,\infty\right)

Polynomial Function: Domain of g(x) = 2x³-1

all real numbers or \left(-\infty,\infty\right)

Radical Function

the inside of the square root for example : a\sqrt{b}x

a is the index

b is the radical

f(x) =\sqrt{x-4}

x - 4 ≥ 0

x ≥ 0

Domain = (4, \infty) or x ≥ 4 the domain cannot be less than 4

h(x) =\sqrt{3x-6}-2

3x - 6 ≥ 0

x ≥ 2

Domain = (2, \infty) or x ≥ 2 the domain cannot be less than 2

g(k) =3\sqrt{3k-9}

If the index is odd then the domain will always be all real numbers because there is no restriction.

Domain: \left(-\infty,\infty\right)

f(x) =4\sqrt{5x-12}

5x - 12 ≥ 0

x = \frac{12}{5}

Domain = [\frac{12}{5} \frac{12}{5} , \infty)

Rational Function

Fraction where numerator and denominator are both polynomials. A fraction cannot have a zero denominator. So whenever you have a fraction, set the denominator NOT EQUAL to zero. x ≠ 0

Find the domain of each rational function.

1. f\left(x\right)=\frac{2x-3}{3x+5}

2. f\left(x\right)=\frac{3}{5x-10}

3. f\left(k\right)=\frac{k^2-64}{k^2-k-56}

4. f\left(z\right)=\frac{z^2+9z+20}{z^2-25}

1. x ≠ \frac{-5}{3}

   Domain : \left(-\infty,-\frac53\right)\cup\left(-\frac{5}{3,}\infty\right)

2. x ≠ 2

   Domain : \left(-\infty,2\right)\cup\left(2,\infty\right)

   \left(-\infty,-\frac53\right)\cup\left(-\frac{5}{3,}\infty\right)

3. k ≠ -7 and k ≠ 8

   Domain : (−∞,−7)∪(−7,8) ∪ (8,∞) (−∞,−7) ∪ (−7,8) ∪ (8,∞)

4. x ≠ -5

   Domain : \left(-\infty,-5\right)\cup\left(-5,5)\right.\cup\left(5,\infty\right)

Fractions with Radicals

1. f(x) = \frac{2x}{\sqrt{x+3}}

2. f(x) = \frac{1}{\sqrt{x+2}}

3. f(x) = \frac{\sqrt{x-3}}{x-7}

4. g(x)= \frac{\sqrt{x-3}}{\sqrt{x-2}}

5. h(x) = \frac{2x+7}{\sqrt{x^2+3x-28}}

6. h(x) = \frac{x+3}{\sqrt{x^2-13x+40}}

1. x+3 > 0

   D: x>-3

   [-3, ♾)

2. x+2>0

   D: x>-2

   [-2, ♾)

3. x ≥ 3

   D: [3,7) U (7, ♾ )

4. x≥ 3

   x>2

5. x< -7 and x>4

   x=4 x=-7

6. x=5 x=8

   x<5 and x>8

1. x ≠ 0

2. x ≠ 0

3. x ≠ 0

4. x ≠ 0