Composition conditions and a hidden constant case

Functions f:x↦(x2−6x−6)/(x+1) f:x\mapsto(x^2-6x-6)/(x+1), x∈Rx\in\mathbb R, x≠−1 x\ne-1; g:x↦(ax+1)/(x+b) g:x\mapsto(ax+1)/(x+b), x∈Rx\in\mathbb R, x≠−bx\ne-b, where a,ba, b are constants.

(i) Sketch y=f(x) y=f(x), giving coordinates of turning points and equations of asymptotes. Write down the range of ff. marks

(ii) Find the value of aa and range of bb such that both fgfg and gfgf exist. marks

(iii) Find g−1(x) g^{-1}(x). Given g−1(x)=g(x) g^{-1}(x)=g(x) for all real xx, x≠−bx\ne-b, find bb in terms of aa. marks

[2019/TJC/Promo/9]

Total marks: 10

Solution

(i)

Sketching rational functions

f(x)=x2−6x−6x+1f(x)=\dfrac{x^2-6x-6}{x+1} for x≠−1x\ne-1. Sketch y=f(x)y=f(x) giving the coordinates of the turning points and the equations of asymptotes. Write down the range of ff.

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Hint

  1. Divide to get f(x)=x−7+1x+1f(x)=x-7+\frac1{x+1}.

  2. Read off the asymptotes.

  3. Differentiate for the turning points.

Rational graph and its two branches
Rational graph and its two branchesScroll across to see the full diagram.

Answer · (i)

Asymptotes x=−1,y=x−7x=-1,y=x-7; maximum (−2,−10)(-2,-10), minimum (0,−6)(0,-6); range (−∞,−10]∪[−6,∞)(-\infty,-10]\cup[-6,\infty).

Explore the idea

Two branches with an oblique asymptote

Switch on the asymptotes and the turning points.

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

  • F
  • Asymptotes
Ideas to try
Is any yy between −10-10 and −6-6 ever reached?
(ii)

Existence of composites

g(x)=ax+1x+bg(x)=\dfrac{ax+1}{x+b} for x≠−bx\ne-b. Find the value of aa and the range of bb such that both fgfg and gfgf exist.

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Hint

  1. Write g(x)=a+1−abx+bg(x)=a+\frac{1-ab}{x+b} so that Rg=R∖{a}R_g=\mathbb R\setminus\{a\}.

  2. fgfg exists iff Rg⊆DfR_g\subseteq D_f.

  3. gfgf exists iff Rf⊆DgR_f\subseteq D_g.

Answer · (ii)

6<b<106<b<10 and either a=−1a=-1 or a=1/ba=1/b.

The a=1/b constant family satisfies literal (ii). If invertibility in (iii) is intended throughout, retain only a=−1.

Explore the idea

Which values does f never take?

Slide y = −b. For −10 < −b < −6 the line misses the graph, so g is never asked to evaluate at −b.

Height of the horizontal linek = -8
Try an example
Test boundary heights

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

Number of distinct intersections
0

No intersection in the gap. This gives the source condition 6 < b < 10.

There are no isolated intersections at this value. A touching point counts once; hollow endpoints are excluded.

  • F
  • Y = −b
  • Inputs where f takes the value −b
  • Intersection selected in the coordinate table
Key idea

The line must stay inside the gap. The turning-point values −10-10 and −6-6 are not allowed, because ff reaches them.

Ideas to try
  1. Find the largest and smallest allowed values of −b-b.
  2. What happens at −b=−10-b=-10 and −b=−6-b=-6 exactly?
(iii)

Self-inverse functions

Find g−1(x)g^{-1}(x). Given that g−1(x)=g(x)g^{-1}(x)=g(x) for all x≠−bx\ne-b, find bb in terms of aa.

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Hint

  1. Solve y=ax+1x+by=\frac{ax+1}{x+b} for xx.

  2. Compare with gg.

Answer · (iii)

g−1(x)=(1−bx)/(x−a)g^{-1}(x)=(1-bx)/(x-a), x≠ax\ne a; self-inverse requires b=−ab=-a.

Inverse exists only when ab≠1.

Explore the idea

An example: a = −1, b = 1

g(x) = (1 − x)/(1 + x) satisfies b = −a. Its graph is symmetric about y = x, so it equals its own inverse. One branch is shown.

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, 0)
Q = (f(x), x)
(0, 1)
on the inverse
Domain of f
x>−1x>-1 (right branch)
Range of f = domain of f⁻¹
y>−1y>-1
Midpoint M on y = x
(0.5, 0.5)
  • G
  • G⁻¹, the same curve
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

b=−ab=-a makes the asymptotes x=−b=ax=-b=a and y=ay=a swap into each other under reflection in y=xy=x, so the curve maps onto itself.

Ideas to try
Move xx and compare PP and QQ: is QQ on the same curve?