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

If this part uses an earlier result, revisit (i) Sketching rational functions or open all working together.

(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.

Explore this idea

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

Open interactive exploration

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?