Porisma

Worked solution

Check whole-domain composition before finding its range

The functions ff and gg are defined by

f:x↦sin⁡(x+π4)f:x\mapsto\sin\left(x+\dfrac\pi4\right), x∈Rx\in\mathbb R, −π2≤x≤π2-\dfrac\pi2\leq x\leq\dfrac\pi2,

g:x↦ln⁡(2−x)g:x\mapsto\ln(2-x), x∈Rx\in\mathbb R, x<2x<2.

Explain whether the following composite function exist

(a) gfgf

(b) fgfg

If the composite function exists, determine the rule, domain and range. marks

[2013/CJC/JC1CA1/3(iv)(modified)]

CJC · Review Problems · Question 2

Answer

(a) gf(x)=ln⁡(2−sin⁡(x+π/4))gf(x)=\ln(2-\sin(x+\pi/4)), domain [−π/2,π/2][-\pi/2,\pi/2], range [0,ln⁡(2+1/2)]≈[0,0.996][0,\ln(2+1/\sqrt2)]\approx[0,0.996]. (b) fgfg does not exist on the full stated inner domain.

(a)

The input to sine ranges over [−π/4,3π/4][-\pi/4,3\pi/4], so Rf=[−1/2,1]R_f=[-1/\sqrt2,1]. Since this lies inside Dg=(−∞,2)D_g=(-\infty,2), gfgf exists on all of DfD_f. Its rule is gf(x)=ln⁡(2−sin⁡(x+π/4))gf(x)=\ln\left(2-\sin(x+\pi/4)\right) and Dgf=[−π/2,π/2]D_{gf}=[-\pi/2,\pi/2]. Because ln⁡(2−t)\ln(2-t) decreases with t<2t<2, its range is [0,ln⁡(2+1/2)][0,\ln(2+1/\sqrt2)], or [0,0.996][0,0.996] in the printed three-significant-figure form.

How did this part go?

(b)

The range Rg=RR_g=\mathbb R is not a subset of Df=[−π/2,π/2]D_f=[-\pi/2,\pi/2]. Thus fgfg does not exist as a composite on the full stated domain of gg. A deliberately restricted inner domain would define a different function; no such restriction is requested here.

How did this part go?

Common mistakes

The source uses the whole-domain convention for existence; a formula that works for some inputs is insufficient.

Solution notes

The source upper range endpoint 0.996 is a rounded value; the exact logarithmic endpoint is retained alongside it.

Read the hint