Functions And Graphs

Question 1

Given that f( x ) = | 4x - 8 | and g( x ) = x + 1

Solve the equation f( x ) = 12

Answer 1

• | 4x - 8 | = 12
• ( We take the positive and negative of this equation )
• 4x - 8 = 12
• x = 5
• • 4x + 8 = 12
• x = - 1

Question 2

Given that f( x ) = | 4x - 8 | and g( x ) = x + 1

Solve f( x ) = g( x )

Answer 2

• | 4x + 8 | = x + 1
• 4x - 8 = x + 1
• x = 3
• • 4x + 8 = x + 1
• x = 7 / 5

Question 3

For each of the following functions:

a ) f : x - > 5x - 2, domain { x = 1, 2, 3, 4, 5 }

b ) g : x - > 3 - x, domain { x = 0, 1, 2, 3 }

c ) h : x - > 4 cos x, domain { - 180 <= x <= 180 }

i ) State whether the function is one-to-one or many-to-one
ii ) Find the range of the function

Answer 3

ai )

• Each value of f( x ) can only arise from one value of x
• Therefore the function is one-to-one

aii )

• ( To find the range, substitute the values of x which has been given in the equation )
• Range = 3, 8, 13, 18, 23

bi )

• Each values of g( x ) can only arise from one value of x
• Therefore the function is one-to-one

bii )

• ( To find the range, substitute the values of x which has been given in the equation )
• Range = 3, 2, 1, 0

ci )

• Value of h( x ) can arise from both x and - x
• ( E.g h( 90 ) = 0 and h( - 90 ) = 0 )
• Therefore the function is many-to-one

cii )

• ( To find the range, substitute the values of x which has been given in the equation )
• Range = - 4 <= h( x ) <= 4

Question 4

The function f is defined by f( x ) = x^2 + 3 with the domain x E R, x => 0

a ) Find an expression for f^- 1 ( x )

b ) State the domain of f^- 1 ( x )

Answer 4

a )

• f( x ) = x^2 + 3
• y = x^2 + 3
• ( Find x )
• y - 3 = x^2
• x = Root y - 3
• ( Replace y with x )
• f^- 1 ( x ) = Root x - 3

b )

• When f^- 1 ( x ) = 0
• Root x - 3 = 0
• Root x = Root 3
• x = 3
• Domain = x E R, x => 3