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Worksheet: Finding the Inverse of a Function from Its Graph

Q1:

Liam is looking for an inverse to 𝑓 ( π‘₯ ) = βˆ’ 2 βˆ’ ( π‘₯ βˆ’ 4 ) 2 . He starts with the parabola 𝑦 = βˆ’ 2 βˆ’ ( π‘₯ βˆ’ 4 ) 2 . He then reflects this in the line 𝑦 = π‘₯ to get the shown parabola π‘₯ = βˆ’ 2 βˆ’ ( 𝑦 βˆ’ 4 ) 2 .

Complete Liam’s work by determining the inverse 𝑓 βˆ’ 1 whose graph is the given solid curve.

  • A 𝑓 ( π‘₯ ) = 4 βˆ’ √ π‘₯ βˆ’ 2 βˆ’ 1
  • B 𝑓 ( π‘₯ ) = 4 + √ βˆ’ π‘₯ βˆ’ 2 βˆ’ 1
  • C 𝑓 ( π‘₯ ) = 4 βˆ’ √ π‘₯ + 2 βˆ’ 1
  • D 𝑓 ( π‘₯ ) = 4 βˆ’ √ βˆ’ π‘₯ βˆ’ 2 βˆ’ 1
  • E 𝑓 ( π‘₯ ) = 4 + √ π‘₯ βˆ’ 2 βˆ’ 1

Q2:

The following graph is of the function 𝑓 ( π‘₯ ) = 6 π‘₯ + 8 π‘₯ + 1 2 , with its maximum at ο€Ό 1 3 , 9  , minimum at ( βˆ’ 3 , βˆ’ 1 ) , and zero at βˆ’ 4 3 labeled.

Find an expression for the inverse function 𝑓 βˆ’ 1 when 𝑓 is restricted to the interval βˆ’ 4 3 < π‘₯ ≀ 1 3 .

  • A 𝑓 ( π‘₯ ) = 3 + √ βˆ’ π‘₯ + 8 π‘₯ + 9 π‘₯ βˆ’ 1 2
  • B 𝑓 ( π‘₯ ) = π‘₯ + 1 6 π‘₯ + 8 βˆ’ 1 2
  • C 𝑓 ( π‘₯ ) = 3 + √ βˆ’ π‘₯ + 8 π‘₯ + 9 βˆ’ 1 2
  • D 𝑓 ( π‘₯ ) = 3 βˆ’ √ βˆ’ π‘₯ + 8 π‘₯ + 9 π‘₯ βˆ’ 1 2
  • E 𝑓 ( π‘₯ ) = 3 βˆ’ √ βˆ’ π‘₯ + 8 π‘₯ + 9 βˆ’ 1 2

What is the domain of 𝑓 βˆ’ 1 in this case?

  • A 0 < π‘₯ ≀ 9
  • B βˆ’ 1 ≀ π‘₯ ≀ 9
  • C βˆ’ 1 < π‘₯ ≀ 9
  • D βˆ’ 4 3 < π‘₯ ≀ 1 3
  • E 0 ≀ π‘₯ ≀ 9

Q3:

The following graph is of the function 𝑓 ( π‘₯ ) = 6 π‘₯ + 8 π‘₯ + 1 2 , with its maximum at ο€Ό 1 3 , 9  , minimum at ( βˆ’ 3 , βˆ’ 1 ) , and zero at βˆ’ 4 3 marked.

Find an expression for the inverse function 𝑓 βˆ’ 1 when 𝑓 is restricted to the interval π‘₯ β‰₯ 1 3 .

  • A 𝑓 ( π‘₯ ) = 3 βˆ’ √ βˆ’ π‘₯ + 8 π‘₯ + 9 π‘₯ βˆ’ 1 2
  • B 𝑓 ( π‘₯ ) = π‘₯ + 1 6 π‘₯ + 8 βˆ’ 1 2
  • C 𝑓 ( π‘₯ ) = 3 + √ βˆ’ π‘₯ + 8 π‘₯ + 9 βˆ’ 1 2
  • D 𝑓 ( π‘₯ ) = 3 + √ βˆ’ π‘₯ + 8 π‘₯ + 9 π‘₯ βˆ’ 1 2
  • E 𝑓 ( π‘₯ ) = 3 βˆ’ √ βˆ’ π‘₯ + 8 π‘₯ + 9 βˆ’ 1 2

What is the domain of 𝑓 βˆ’ 1 in this case?

  • A 0 < π‘₯ ≀ 9
  • B βˆ’ 1 ≀ π‘₯ < 9
  • C 0 ≀ π‘₯ ≀ 9
  • D π‘₯ β‰₯ 1 3
  • E βˆ’ 1 < π‘₯ ≀ 9

Q4:

The graphs of 𝑓 ( π‘₯ ) = π‘₯ + 𝑏  and its inverse 𝑓 ( π‘₯ )   intersect at three points, one of which is ο€Ό 4 5 , 4 5  .

Determine the value of 𝑏 .

  • A 6 4 1 2 5
  • B 1 6 4 1 2 5
  • C 4 5
  • D 3 6 1 2 5
  • E βˆ’ 4 5

Find the π‘₯ -coordinate of the point 𝐴 marked on the figure.

  • A 3 6 1 2 5
  • B 4 5
  • C 6 4 1 2 5
  • D 1 6 4 1 2 5
  • E βˆ’ 4 5

Find the π‘₯ -coordinate of the point 𝐡 marked on the figure.

  • A 8 2 5
  • B βˆ’ 2 βˆ’ √ 1 3 5
  • C √ 1 3 βˆ’ 2 5
  • D 3 6 1 2 5
  • E 1 3

Q5:

Determine whether the inverse of the represented function is a function or not.

  • Anot a function
  • Ba function

Q6:

The following graph is of the function , with its maximum at , minimum at , and zero at marked.

Find an expression for the inverse function when is restricted to the interval .

  • A
  • B
  • C
  • D
  • E

What is the domain of in this case?

  • A
  • B
  • C
  • D
  • E

Q7:

The following is the graph of 𝑓 ( π‘₯ ) = 2 π‘₯ βˆ’ 1 .

Which one is the graph of the inverse function 𝑓 ( π‘₯ ) βˆ’ 1 ?

  • A (c)
  • B (b)
  • C (a)

Q8:

In the given figure, the green points represent the function 𝑓 ( π‘₯ ) . Do the blue points represent 𝑓 ( π‘₯ ) βˆ’ 1 ?

  • AYes
  • BNo

Q9:

Consider the two following figures.

The first figure shows the graph of 𝑓 ( π‘₯ ) = π‘₯  and a tangent to the graph with gradient 1. This tangent meets the graph at a point with π‘₯ -coordinate 1 √ 3 .

The second figure shows the graphs of 𝑔 ( π‘₯ ) = π‘₯ + 𝑏  and its inverse 𝑔 ( π‘₯ ) = ( π‘₯ βˆ’ 𝑏 )     . The graphs cross in the third quadrant and touch in the first quadrant.

What is the value of 𝑏 ?

  • A 𝑏 = βˆ’ 2 √ 3 9
  • B 𝑏 = 4 √ 3 9
  • C 𝑏 = βˆ’ 4 √ 3 9
  • D 𝑏 = 2 √ 3 9
  • E 𝑏 = βˆ’ 1 √ 3

What are the π‘₯ -coordinates of the two points of intersection of the graphs in the second figure?

  • A 1 √ 3 and βˆ’ 2 √ 3
  • B 1 √ 3 and βˆ’ 5 √ 3
  • C 1 √ 3 and βˆ’ √ 3 9
  • D √ 3 and βˆ’ 1 √ 3
  • E √ 3 and βˆ’ 5 √ 3