When can two functions compose?

The functions are f(x)=x−2f(x)=\sqrt{x-2} for x≥2x\ge2, and g(x)=3x2−6x+5g(x)=3x^2-6x+5 for 0≤x≤40\le x\le4, with real inputs.

(i) Sketch y=f(x) and state its range. marks

(ii) Find the inverse of f and its domain. marks

(iii) Show that fg exists. marks

(iv) Find fg(x) and its range. marks

(v) With g now defined on all real numbers, describe fully a sequence of transformations taking y=g(x) to y=x2−1y=x^2-1. marks

[2023/MI/Promo/PU2/I/8]

Total marks: 12

Solution

If this part uses an earlier result, revisit (i) Range from a graph or open all working together.

(ii)

Inverse functions

Find the inverse of ff and state its domain.

Explore this idea

Hint

  1. Write y=x−2y=\sqrt{x-2} and make xx the subject.

  2. Swap the names xx and yy.

  3. The domain of f−1f^{-1} is the range of ff.

Answer · (ii)

f−1(x)=x2+2f^{-1}(x)=x^2+2, x≥0x\ge0.

Explore the idea

Open interactive exploration

f and f⁻¹ are mirror images in y = x

Drag the input. P is the point (x, f(x)) and Q is its mirror image (f(x), x), which lies on the inverse.

Point on f: input xx = 4.00
Try an example

Reflection swaps the coordinates. The segment PQ is perpendicular to y = x, and its blue midpoint M lies on that line.

P = (x, f(x))
(4, 1.41421)
Q = (f(x), x)
(1.41421, 4)
on the inverse
Domain of f
x≥2x\ge2
Range of f = domain of f⁻¹
y≥0y\ge0
Midpoint M on y = x
(2.70711, 2.70711)
  • F(x) = √(x − 2)
  • F⁻¹(x) = x² + 2, for x ≥ 0
  • Mirror y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

Reflecting in y=xy=x swaps inputs and outputs: the domain of ff becomes the range of f−1f^{-1}, and vice versa.

Ideas to try
  1. Move xx to 2. Where is QQ? Why is the inverse's graph starting at (0,2)(0,2)?
  2. Why is the inverse only drawn for x≥0x\ge0?