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

(a) one-one

One-one functions

f(x)=(x−2)(x−4)f(x)=(x-2)(x-4) for x<4x<4 and g(x)=xx+3g(x)=\dfrac x{x+3} for x≠−3x\ne-3. Determine whether ff is one-one.

Explore this idea

Hint

  1. Look for two inputs with the same output: the vertex is at x=3x=3 and the domain extends beyond it.

Answer · (a) one-one

Not one-one: f(5/2)=f(7/2)=−3/4f(5/2)=f(7/2)=-3/4.

Explore the idea

The horizontal line test on (x − 2)(x − 4), x < 4

Between −1 and 0 the line cuts the graph twice. At or above 0, only the left intersection lies in x < 4.

Height of the horizontal linek = -0.75
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct intersections
2

Two intersections, both within x < 4. This proves the function is not one-one.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
2.5-0.75
3.5-0.75
  • F
  • Y = k
  • Intersections
  • Intersection selected in the coordinate table
Key idea

The domain x<4x<4 takes in both sides of the vertex, so the graph turns back on itself.

Ideas to try
Set k=−0.75k=-0.75: which two inputs do you find?
(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

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?
(b) inverse

Inverse functions

h(x)=a(1+e−x)h(x)=a(1+e^{-x}) for x>0x>0, where a>0a>0. Find h−1(x)h^{-1}(x) and its domain.

Explore this idea

Hint

  1. Write y=a(1+e−x)y=a(1+e^{-x}) and make xx the subject.

  2. Dh−1=RhD_{h^{-1}}=R_h.

Answer · (b) inverse

h−1(x)=−ln⁡(x/a−1)h^{-1}(x)=-\ln(x/a-1).

If ℝ₀⁺ means x≥0: inverse domain (a,2a]. If intended x>0 as solution uses: (a,2a). Endpoint ambiguity remains explicit.

Explore the idea

h and its inverse for a = 2

Each curve has an asymptote that is the mirror of the other's: y = a for h, and x = a for h⁻¹.

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, 2.73576)
Q = (f(x), x)
(2.73576, 1)
on the inverse
Domain of f
x≥0x\ge0
Range of f = domain of f⁻¹
a<y≤2aa<y\le2a
Midpoint M on y = x
(1.86788, 1.86788)
  • H(x) = 2(1 + e⁻ˣ)
  • H⁻¹
  • Y = x, where h and h⁻¹ meet
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

The two curves meet exactly on y=xy=x, at the point (b,b)(b,b) used in the area question.

Ideas to try
Where do hh and h−1h^{-1} meet? What is that xx called in the question?
(b) area

Area using symmetry

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

Explore this idea

Hint

  1. The two curves are mirror images in y=xy=x, meeting at (b,b)(b,b).

  2. The area is twice the area between hh and y=xy=x on [0,b][0,b].

  3. That area is II minus the triangle under y=xy=x.

Answer · (b) area

2I−b22I-b^2.

Explore the idea

Two mirror-image regions meeting at (b, b)

Each half is the area between a curve and y = x. Doubling gives the whole enclosed area.

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

  • H
  • H⁻¹
  • Y = x
Ideas to try
Which region is I? Which part of it must be removed before doubling?