Invert a special point separately

For all real x, f(x)=cos⁡xf(x)=\cos x. For −1<x≤1-1<x\le1, g(x)=ln⁡(1/(1+x))g(x)=\ln(1/(1+x)). Define h by the same logarithmic rule for −1<x<1-1<x<1, but set h(1)=−1.

(i) Show fg exists. marks

(ii) Expand fg through the cubic term. marks

(iii) State h−1(−1)h^{-1}(-1), then define the piecewise inverse as p(x) for x>a and c for x=−1, finding a,c and p. marks

[2010/MJC/Prelim/I/10]

Total marks: 10

Solution

(i)

Existence of composites

f(x)=cos⁡xf(x)=\cos x for all real xx and g(x)=ln⁡11+xg(x)=\ln\dfrac1{1+x} for −1<x≤1-1<x\le1. Show that fgfg exists.

Explore this idea

Hint

  1. Find RgR_g from the graph of −ln⁡(1+x)-\ln(1+x).

  2. Compare with Df=RD_f=\mathbb R.

Answer · (i)

fgfg exists: Rg=[−ln⁡2,∞)⊆R=DfR_g=[-\ln2,\infty)\subseteq\mathbb R=D_f.

Explore the idea

g falls from +∞ to −ln 2

Switch on the asymptote and the closed end-point.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • G
  • Asymptote
Ideas to try
Is any yy below −ln⁡2-\ln2 ever reached?
(ii)

Maclaurin series of composites

Expand fg(x)fg(x) in ascending powers of xx up to and including x3x^3.

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Hint

  1. fg(x)=cos⁡(ln⁡(1+x))fg(x)=\cos\left(\ln(1+x)\right) because cos⁡\cos is even.

  2. Expand ln⁡(1+x)\ln(1+x) to the cubic term.

  3. Substitute into cos⁡u=1−u22+⋯\cos u=1-\frac{u^2}2+\cdots and keep terms up to x3x^3.

Answer · (ii)

fg(x)=1−x2/2+x3/2+O(x4)fg(x)=1-x^2/2+x^3/2+O(x^4).

Maclaurin expansion near x=0.

Explore the idea

The cubic matches the composite near 0

The series 1 − x²/2 + x³/2 sits on top of cos(ln(1 + x)) near x = 0 and drifts away further out.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Composite
  • Cubic approximation
Ideas to try
Compare the two curves at x=0.5x=0.5, then at x=1x=1.
(iii) value

Isolated points

hh has the same rule as gg for −1<x<1-1<x<1, but h(1)=−1h(1)=-1. State h−1(−1)h^{-1}(-1).

Hint

  1. h−1(−1)h^{-1}(-1) is the input that hh maps to −1-1.

Answer · (iii) value

h−1(−1)=1h^{-1}(-1)=1.

(iii) inverse

Piecewise inverse

The inverse of hh is h−1(x)=p(x)h^{-1}(x)=p(x) for x>ax>a and cc for x=−1x=-1. Find aa, cc and p(x)p(x).

Explore this idea

Hint

  1. On −1<x<1-1<x<1, h=g=−ln⁡(1+x)h=g=-\ln(1+x): solve y=−ln⁡(1+x)y=-\ln(1+x).

  2. The range of this branch fixes where pp applies.

  3. The isolated point gives the constant cc.

Answer · (iii) inverse

p(x)=e−x−1p(x)=e^{-x}-1, a=−ln⁡2a=-\ln2, c=1c=1.

Continuous inverse branch x>−ln2; isolated inverse input x=−1.

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The main branch and its inverse

g(x) = −ln(1 + x) for −1 < x < 1 and its inverse eˣ-based branch. The extra point (1, −1) maps to the isolated point (−1, 1).

Point on f: input xx = 0.00
Reset the controls

P and Q coincide on y = x. This input is a fixed point.

P = (x, f(x))
(0, 0)
Q = (f(x), x)
(0, 0)
on the inverse
Domain of f
−1<x<1-1<x<1
Range of f = domain of f⁻¹
y>−ln⁡2y>-\ln2
Midpoint M on y = x
(0, 0)
  • H on −1 < x < 1
  • P(x) = e⁻ˣ − 1 for x > −ln 2
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

The isolated point h(1)=−1h(1)=-1 produces its own isolated inverse value h−1(−1)=1h^{-1}(-1)=1, so the inverse is defined in two pieces.

Ideas to try
Where does the inverse begin on the left? Why does it not include x=−ln⁡2x=-\ln2?