A logarithmic involution

(a) The curve y=f(x) y=f(x) has horizontal asymptote y=ky=k and cuts the axes at (a,0)(a, 0) and (0,b)(0, b), where a,b,ka, b, k are non-zero constants. Given that f−1 f^{-1} exists, state if possible the coordinates where the following curves cut the axes and their asymptotes: (i) y=f(2x−3) y=f(2x-3), (ii) y=f−1(x) y=f^{-1}(x). marks

(b) g:x↦ln⁡(ex+5ex−1) g:x\mapsto\ln\left(\frac{e^x+5}{e^x-1}\right) for x∈Rx\in\mathbb R, x>0 x>0.

(i) Find g−1(x) g^{-1}(x) and state its domain. marks

The function h:x↦1+9−(x−2)2 h:x\mapsto1+\sqrt{9-(x-2)^2} for x∈Rx\in\mathbb R, −1≤x≤5-1\le x\le5.

(ii) Find the exact solutions of g2021h(x)=ln⁡2 g^{2021}h(x)=\ln2 in simplest form. marks

[2021/NYJC/Prelim/I/5]

Total marks: 8

Solution

If this part uses an earlier result, revisit (a)ii Inverse functions or open all working together.

(b)i

Self-inverse functions

g(x)=ln⁡ex+5ex−1g(x)=\ln\dfrac{e^x+5}{e^x-1} for x>0x>0. Find g−1(x)g^{-1}(x) and state its domain.

Explore this idea

Hint

  1. Write y=g(x)y=g(x), exponentiate, and make exe^x the subject.

  2. Compare with gg.

Answer · (b)i

g−1(x)=ln⁡((ex+5)/(ex−1))=g(x)g^{-1}(x)=\ln((e^x+5)/(e^x-1))=g(x), x>0x>0.

Explore the idea

Open interactive exploration

g is its own mirror image

The graph is symmetric about y = x, so reflecting it in y = x gives the same curve.

Point on f: input xx = 1.00
Reset the controls

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))
(1, 1.50227)
Q = (f(x), x)
(1.50227, 1)
on the inverse
Domain of f
x>0x>0
Range of f = domain of f⁻¹
y>0y>0
Midpoint M on y = x
(1.25113, 1.25113)
  • G
  • G⁻¹, the same curve
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

A curve that is symmetric about y=xy=x is self-inverse.

Ideas to try
Move xx: is QQ always on the same curve as PP?