restricted quadratic inverse

Practise restricted quadratic inverse with an original VectorPaths question, worked explanation and related VCE Methods practice.

3 practice marks

Original practice question

Let $f(x)=2 \left(x - 1\right)^{2} - 3$ for $x\geq 1$. Find the rule and domain of $f^{-1}$.

Worked explanation

• Reverse the function

  Solve $y=2 \left(x - 1\right)^{2} - 3$: $(x-(1))^2=(y-(-3))/2$.

• Choose the valid branch

  The restriction $x\geq 1$ requires $x-(1)\geq0$, so take the positive square root.

• State rule and domain

  $f^{-1}(x)=\sqrt{\frac{x}{2} + \frac{3}{2}} + 1$. Its domain is the range of $f$: $[-3,\infty)$.

$f^{-1}(x)=\sqrt{\frac{x}{2} + \frac{3}{2}} + 1$, with domain $[-3,\infty)$.

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