Function Composition and Inverses: math test solutions, Grade 12 – download the PDF

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Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

Sommaire

(a) \((f\circ g)(x)=(3x+1)^2-4(3x+1)=9x^2+6x+1-12x-4=9x^2-6x-3\). (1.5 pts)

(b) \((g\circ f)(x)=3(x^2-4x)+1=3x^2-12x+1\). (1 pt)

(d) \(9x^2-6x-3=0\) gives \(3x^2-2x-1=0\), that is \((3x+1)(x-1)=0\). So \(x=1\) or \(x=-\dfrac13\). (1 pt)

\((f\circ g)(x)=\sqrt{5-x^2}\). We need \(5-x^2\ge0\), so the domain is \([-\sqrt5,\sqrt5]\). (1.5 pts)

\((g\circ f)(x)=6-(\sqrt{x-1})^2=7-x\), with domain \([1,\infty)\) because \(f\) needs \(x\ge1\). (1.5 pts)

(a) \(y(x-1)=2x+5\) gives \(xy-y=2x+5\) and \(x(y-2)=y+5\). So \(f^{-1}(x)=\dfrac{x+5}{x-2}\). (2.5 pts)

(b) \(f(f^{-1}(x))=\dfrac{\frac{2(x+5)}{x-2}+5}{\frac{x+5}{x-2}-1}=\dfrac{7x/(x-2)}{7/(x-2)}=x\). (1 pt)

\(f(x)=(x-3)^2-5\), range \([-5,\infty)\). (1 pt)

\(y=(x-3)^2-5\) with \(x\ge3\) gives \(x=3+\sqrt{y+5}\). So \(f^{-1}(x)=3+\sqrt{x+5}\), domain \([-5,\infty)\). (1 pt)

\(f^{-1}(11)=3+4=7\) and \(f(7)=49-42+4=11\). (1 pt)

(a) Shift left 3, stretch vertically by 2, shift down 4. Domain \([-3,\infty)\), range \([-4,\infty)\). (1.5 pts)

(b) \(g(x)=0\): \(\sqrt{x+3}=2\), so \(x=1\). y-intercept: \(g(0)=2\sqrt3-4\approx-0.54\). (1.5 pts)

(a) \(f(-x)=-x^5-3x=-f(x)\): odd. \(g(-x)=x^4-|x|=g(x)\): even. \(h(1)=4\) and \(h(-1)=0\): neither. (1.5 pts)

(b) If \(x\lt1\), \(|x-1|=1-x\), so \(k(x)=1\). If \(x\ge1\), \(k(x)=(x-1)+x=2x-1\). (1.5 pts)

Test yourself: quick challenge for Grade 12