One-to-one restriction and inverse

Practise one-to-one restriction and inverse with an original VectorPaths question, worked explanation and related VCE Methods practice.

4 practice marks

Original practice question

Let $f(x)=(x-3)^2+2$ for $x\geq3$. (a) Explain why an inverse exists on this domain. [1 marks] (b) Find the inverse rule and its domain. [3 marks]

Worked explanation

• Part (a)

  The square $(x-3)^2$ increases strictly when $x\geq3$, so no two inputs give the same output.

• Part (b)

  $y-2=(x-3)^2$. The restriction gives $x-3\geq0$, so $x=3+\sqrt{y-2}$. Swap the variables; the inverse domain is the original range $[2,\infty)$.

(a) f is strictly increasing.; (b) $f^{-1}(x)=3+\sqrt{x-2}$, $x\geq2$.

Original VectorPaths practice; not an official VCAA question or marking rubric.