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 $f(x)=3 \left(x + 2\right)^{2} + 4$ for $x\geq -2$. Find the rule and domain of $f^{-1}$.

Worked explanation

• Reverse the function

  Solve $y=3 \left(x + 2\right)^{2} + 4$: $(x-(-2))^2=(y-(4))/3$.

• Choose the valid branch

  The restriction $x\geq -2$ requires $x-(-2)\geq0$, so take the positive square root.

• State rule and domain

  $f^{-1}(x)=\sqrt{\frac{x}{3} - \frac{4}{3}} - 2$. Its domain is the range of $f$: $[4,\infty)$.

$f^{-1}(x)=\sqrt{\frac{x}{3} - \frac{4}{3}} - 2$, with domain $[4,\infty)$.

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