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Function Notation
Function Fun
Solving Functions
Functions for Life
Fearless Functions
100

f(x) = x + 5, 

when x=10

x = 15

100

f(x) = x^3 + 4 

g(x) = 2x - 12 

h(x) = x2 - 2x + 3

f(1)

5

100

f(x) = 2x + 1 

Solve when f(x) = 11

x = 5

100

Evaluate the value of a car after 5 years. 

V(t) = 20,000(0.82)^t 

V(5) =

The value of the car is $7414.80 after 5 years.

100

Solve f(x) = (x-3)2 +5 

when f(7)

21

200

f(x) = 3x - 2,

when x=5

x = 13

200

f(x) = x^3 + 4

g(x) = 2x - 12

h(x) = x2 - 2x + 3

g(-2)

-16

200

f(x) = -4x - 2 

Solve when f(x) = -22

x = 5

200

Evaluate the value of a car after 10 years. 

V(t) = 30,000(0.82)^t 

V(10) =

the value of the car is $4123.44 after 10 years.

200

Solve f(x) = (x-3)2 +4 

when f(10)

53

300

f(x) = 2x - 6, 

when x=-10

x = -26

300

f(x) = x^3 + 4 

g(x) = 2x - 12 

h(x) = x2 - 2x + 3

f(-2)

-4

300

f(x) = 5x + 1 

Solve when f(x) = 26

5

300

Evaluate the value of a classic car after 5 years. 

V(t) = 20,000(1.08)^t 

V(5) =

The value of a classic car appreciates to $29,386.56 after 5 years.

300

Solve f(x) = (x-2)3 

when f(6)

64

400

g(x) = x2 + 5, 

when x=2

x = 9

400

f(x) = x^3 + 4 

g(x) = 2x - 12 

h(x) = x2 - 2x + 3

g(h(3))

0

400

g(x) = x2 + 1

Solve when g(x) = 10

x=3

400

Evaluate the value of a classic car after 10 years. 

V(t) = 20,000(1.08)^t 

V(10) =

The value of a classic car appreciates to $43,178.50 after 5 years.

400

Solve f(x) = (x+5)2 - 6 

when f(1)

30

500

g(x) = x2 - 10,

when x=-5

x = 15

500

f(x) = x^3 + 4 

g(x) = 6x - 12 

h(x) = x2 - 2x + 3

f(g(3))

220

500

g(x) = x3 + 5 

Solve when g(x) = 32

x = 3

500

How much will you have in your bank account in 30 years if you put $100 into it right now, with a 5% interest rate compounded annually? 

A=P(1+r/n)^(nt)

You will have $432.19.

500

Solve f(x) = (x-3)2 + 2 

when f(a)

(a-3)2 + 2