2024 JPJC Functions Assignment 4 Q1

2024 JPJC Functions Assignment 4 Q1

12 marks
Tutorial

The function \(\mathrm{f}\) is given by

\(\mathrm{f}:x\mapsto \frac{4}{1+{{x}^{2}}}\), \(x\in \mathbb{R}\)

  1. Show that \({{\mathrm{f}}^{-1}}\) does not exist.[2]
  2. Given that the largest possible domain for which \(\mathrm{f}\) is a one-one function is the set \(\left\{ x:x\in \mathbb{R},x\le A \right\}\), state the value of \(A\).[1]

In the rest of the question, the domain of \(\mathrm{f}\) is \(x\in \mathbb{R}\), \(x\le A\), where \(A\) is the value found in part (ii).

  1. Find \({{\mathrm{f}}^{-1}}\left( x \right)\) and write down the domain of \({{\mathrm{f}}^{-1}}\).[4]
  2. On the same diagram, sketch the graphs of \(y=\mathrm{f}\left( x \right)\), \(y={{\mathrm{f}}^{-1}}\left( x \right)\) and \(y=\mathrm{f}{{\mathrm{f}}^{-1}}\left( x \right)\).[3]
  3. The composite function \(\mathrm{f}\,\mathrm{h}\) is defined by \(\mathrm{f}\,\mathrm{h}:x\mapsto {{x}^{2}}\), \(0 < x\le 2\). Find \(\mathrm{h}\left( x \right)\) and state its domain.[2]

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Answer:\(f^{-1}\) does not exist; \(A = 0\); \(f^{-1}(x) = -\sqrt{\frac{4}{x} - 1}\), \(D_{f^{-1}} = (0, 4]\); \(h(x) = -\sqrt{\frac{4}{x^2} - 1}\), \(D_h = (0, 2]\)

Need help? Join our JC Math tuition classes.

Learn more