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IMAT Mathematics · Functions

Inverse Functions and Transforming Graphs

An inverse function undoes a function and swaps its domain and range; shifts, stretches and reflections move a known graph without changing its basic shape.

Why this matters

The 2023 paper asked for the inverse of a logarithmic function together with the set of values on which that inverse is defined. Transformations have not been asked yet, but they are on the syllabus and let you sketch any of the standard graphs after a shift or stretch.

Concept 1 of 2: Inverse functions: undoing a function

If ff turns 4 into 9, the inverse turns 9 back into 4. To find it, write y=f(x)y = f(x) and solve for xx: you get the input from the output. Inputs and outputs swap roles, so the domain of the inverse is the range of the original.

Definition

The inverse f−1f^{-1} of a function ff satisfies f−1(f(x))=xf^{-1}(f(x)) = x and f(f−1(y))=yf(f^{-1}(y)) = y.

  • It exists only when ff is one-to-one: no two inputs give the same output (so x2x^2 on all real numbers has no inverse, but on x≥0x \ge 0 it does).
  • Method: write y=f(x)y = f(x), solve for xx in terms of yy. That expression is f−1(y)f^{-1}(y).
  • The domain of f−1f^{-1} is the range of ff, and the other way round.
  • The graph of f−1f^{-1} is the reflection of the graph of ff in the line y=xy = x.
  • exe^x and ln⁡x\ln x are inverses: ln⁡(ex)=x\ln(e^x) = x and eln⁡x=xe^{\ln x} = x for x>0x > 0.

Inverse function

y=f(x)⇒x=f−1(y)y = f(x) \quad\Rightarrow\quad x = f^{-1}(y)

Worked example

Let f(x)=3ln⁡x+2f(x) = 3\ln x + 2 for x>0x > 0. Find f−1f^{-1} and state where it is defined.
Practice this conceptself-check · 4 quick reps

The inverse function is not the reciprocal

f−1(x)f^{-1}(x) means the function that undoes ff; 1f(x)\dfrac{1}{f(x)} is the reciprocal, a different thing. For f(x)=2xf(x) = 2x, f−1(x)=x/2f^{-1}(x) = x/2 but 1/f(x)=1/(2x)1/f(x) = 1/(2x).

Give the inverse with its domain

An option can have the right formula and the wrong domain. The domain of f−1f^{-1} is the range of ff: work out which outputs ff really produces before choosing.

Concept 2 of 2: Transformations of graphs: shifts, stretches and reflections

Changes outside the function act on the output, so they move the graph up or down or stretch it vertically, exactly as written. Changes inside the bracket act on the input, so they work backwards: f(x−3)f(x - 3) reaches each value 3 units later, which moves the graph right, not left.

Definition

Starting from the graph of y=f(x)y = f(x):

  • y=f(x)+ky = f(x) + k: shift up by kk (down if k<0k < 0).
  • y=f(x−h)y = f(x - h): shift right by hh (left if h<0h < 0).
  • y=af(x)y = a f(x): vertical stretch by factor aa.
  • y=f(bx)y = f(bx): horizontal stretch by factor 1/b1/b (squeezed towards the y-axis if b>1b > 1).
  • y=−f(x)y = -f(x): reflection in the x-axis; y=f(−x)y = f(-x): reflection in the y-axis.
  • Shifts move asymptotes and key points with them, so they change domains and ranges.

Stretch, then shift

y=a f(x−h)+ky = a\, f(x - h) + k
  • aavertical stretch factor (a reflection in the x-axis if negative)
  • hhshift to the right
  • kkshift upwards

Worked example

Describe how to obtain y=2(x−3)2+1y = 2(x - 3)^2 + 1 from y=x2y = x^2 and give its vertex. Then find the domain of y=ln⁡(x+2)y = \ln(x + 2) and where it crosses the x-axis.
Practice this conceptself-check · 4 quick reps

f(x minus 3) moves the graph right, not left

A change inside the bracket works against its sign: f(x−3)f(x - 3) is the graph moved 3 units right, f(x+3)f(x + 3) is it moved 3 units left. Outside the bracket the sign means what it says: f(x)−3f(x) - 3 is 3 units down.

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