MathematicsGrade 10–12

Inverse functions: reflecting in y = x

Reflect the graph in y = x and the inverse function appears. It only works if no function value occurs twice.

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What is it about?

The inverse function f−1f^{-1} reverses ff: from f(3)=5f(3) = 5 we get f−1(5)=3f^{-1}(5) = 3. This only works if every value occurs exactly once – every horizontal line meets the graph at most once. That is certainly the case if ff is strictly monotonic. x2x^2 on all of R\mathbb{R} is not invertible (f(−2)=f(2)=4f(-2) = f(2) = 4); restricted to x≥0x \geq 0 its inverse is x\sqrt{x}, restricted to x≤0x \leq 0 it is −x-\sqrt{x}. Monotonicity is sufficient but not necessary: 1x\frac{1}{x} is invertible (it is even its own inverse).

To find f−1f^{-1}: solve y=f(x)y = f(x) for xx, then swap xx and yy. Graphically this is a reflection in y=xy = x: P(a∣b)P(a \mid b) becomes P′(b∣a)P'(b \mid a), and domain and range swap. Writing DfD_f for the domain and RfR_f for the range of ff: Df−1=RfD_{f^{-1}} = R_f, Rf−1=DfR_{f^{-1}} = D_f (German textbooks write WfW_f for the range, from Wertemenge). This is how ln⁡x\ln x arises from exe^x: Df=RD_f = \mathbb{R} and Rf=R+R_f = \mathbb{R}^+, so ln⁡x\ln x is defined only for x>0x > 0.

Reflecting the tangent swaps the legs of the slope triangle, so for b=f(a)b = f(a) and f′(a)≠0f'(a) \neq 0:

(f−1)′(b)=1f′(a),(ln⁡x)′=1eln⁡x=1x.\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}, \qquad (\ln x)' = \frac{1}{e^{\ln x}} = \frac{1}{x}.

If f′(a)=0f'(a) = 0, the inverse has a vertical tangent at (b∣a)(b \mid a), e.g. 0.5x30.5x^3 at 00.

Try it

  1. Square root: drag PP or P′P', then press “Reflect in y = x” and watch the graph flip over, the intervals on the axes and the columns of the table.
  2. Line: press “Working step by step”.
  3. x² on all of ℝ: sweep the horizontal line, then switch to a branch.
  4. eˣ and ln x and the vertical tangent.
  5. 1/x and, as an outlook, restricted sine.

Tasks

Task 1: Invert a linear function

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Find the inverse of f(x)=2x−1f(x) = 2x - 1 and f−1(5)f^{-1}(5).

Show solution

x=y+12x = \frac{y + 1}{2}, so f−1(x)=x+12f^{-1}(x) = \frac{x + 1}{2} and f−1(5)=3f^{-1}(5) = 3; check: f(3)=5f(3) = 5. Show

Task 2: Derivative of the square root

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For f(x)=x2f(x) = x^2, x≥0x \geq 0, and P(1.5∣2.25)P(1.5 \mid 2.25): find the slope of x\sqrt{x} at 2.252.25.

Show solution

f′(1.5)=3f'(1.5) = 3, so (f−1)′(2.25)=13\left(f^{-1}\right)'(2.25) = \frac{1}{3}; check: 122.25=13\frac{1}{2\sqrt{2.25}} = \frac{1}{3}. Show

Task 3: Vertical tangent

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Why is 2x3\sqrt[3]{2x} not differentiable at 00? Find (f−1)′(4)\left(f^{-1}\right)'(4) for f(x)=0.5x3f(x) = 0.5x^3.

Show solution

f′(0)=0f'(0) = 0: the horizontal tangent becomes vertical. f(2)=4f(2) = 4 and f′(2)=6f'(2) = 6, so (f−1)′(4)=16\left(f^{-1}\right)'(4) = \frac{1}{6}. Show

Ideas for the classroom

  • Use the views in order: “Invertible?” → “Reflect in y = x” → “Derivative of f⁻¹” leading to (ln⁡x)′=1x(\ln x)' = \frac{1}{x}.
  • Draw a graph on a transparency and flip it along y=xy = x, then compare with the animation.
  • With show=0 and inv=0 the pictures work as exercises.
  • Common stumbling blocks:
    • confusing f−1(x)f^{-1}(x) with 1f(x)\frac{1}{f(x)};
    • forgetting to swap xx and yy, or forgetting the restricted domain (sign of −x-\sqrt{x});
    • thinking “strictly monotonic” requires f′>0f' > 0 everywhere (0.5x30.5x^3 has f′(0)=0f'(0) = 0);
    • evaluating f′f' at bb instead of aa in (f−1)′(b)=1f′(a)\left(f^{-1}\right)'(b) = \frac{1}{f'(a)}.