Functions And Graphs Flashcards

1
Q

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

Solve the equation f( x ) = 12

A
  • | 4x - 8 | = 12
  • ( We take the positive and negative of this equation )
  • 4x - 8 = 12
  • x = 5
    • 4x + 8 = 12
  • x = - 1
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2
Q

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

Solve f( x ) = g( x )

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

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

A

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
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4
Q

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 )

A

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
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