Compose through a restricted range, then use symmetry for area

(a) Functions f:x↦(x−2)(x−4) f:x\mapsto(x-2)(x-4) for x∈Rx\in\mathbb R, x<4 x<4; g:x↦x/(x+3) g:x\mapsto x/(x+3) for x∈Rx\in\mathbb R, x≠−3 x\ne-3.

Determine if ff is one-one, justifying your answer. mark

Only one of fgfg and gfgf exists. Give a definition (including domain) of the composite that exists, and explain why the other does not exist. Find the range of the composite that exists. marks

(b) h:x↦a(1+e−x) h:x\mapsto a(1+e^{-x}), x∈R0+ x\in\mathbb R_0^+, where aa is positive. Find h−1(x) h^{-1}(x) and its domain. marks

Given that hh and h−1 h^{-1} intersect where x=bx=b, and ∫0bh(x) dx=I\int_0^b h(x)\, dx=I , find the area bounded by h,h−1 h, h^{-1} and the axes in terms of I,bI, b. marks

[2010/TJC/Prelim/II/3]

Total marks: 12

Solution

If this part uses an earlier result, revisit (a) one-one One-one functions or open all working together.

(a) composite

Existence and range of composites

Only one of fgfg and gfgf exists. Define the one that exists (including its domain), explain why the other does not, and find the range of the composite that exists.

Explore this idea

Hint

  1. Compare ranges and domains for both orders.

  2. Compose the one that exists and simplify.

  3. Find its range from the range of ff.

Answer · (a) composite

gf(x)=(x2−6x+8)/(x2−6x+11)gf(x)=(x^2-6x+8)/(x^2-6x+11), x<4x<4; range [−1/2,1)[-1/2,1).

gf exists because Rf avoids −3. fg does not exist because Rg contains inputs≥4.

Explore the idea

Open interactive exploration

gf(x) = (x² − 6x + 8)/(x² − 6x + 11)

The composite has a minimum at x = 3 and approaches 1 on the left.

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

  • Gf
  • Asymptote y = 1
Ideas to try
Read the range off the graph: lowest value, highest value, open or closed?