If \(h(x) = g^{-1}(x) + k\) is a tangent to \(f\), determine the coordinates of the point of contact between \(h\) and \(f\).
About this question
Question 4.2.6 of Mathematics Grade 12 Paper 1, NSC Examination (DBE, November 2020). It tests the Term 1 ATP topic Functions and Inverse Functions.
- TermTerm 1
- ATP topicFunctions and Inverse Functions
- PaperPaper 1
- Marks4 marks for the method (formula and substitution) and the correct final answer
The graphs of \(f(x) = \frac{1}{2}(x+5)^2 - 8\) and \(g(x) = \frac{1}{2}x + \frac{9}{2}\) are sketched.
- A is the turning point of \(f\).
- The axis of symmetry of \(f\) intersects the \(x\)-axis at E and the line \(g\) at \(D(m\,;\,n)\).
- C is the \(y\)-intercept of \(f\) and \(g\).
Diagram: a parabola \(f\) opening upwards with its turning point A below the \(x\)-axis to the left of the \(y\)-axis. The straight line \(g\) has a positive gradient and cuts the parabola at its left \(x\)-intercept and again at C on the positive \(y\)-axis. A dotted vertical line through A (the axis of symmetry) meets the \(x\)-axis at E and the line \(g\) at \(D(m\,;\,n)\), which lies above the \(x\)-axis. O is the origin.
\(h(x) = 2x - 9 + k\) has gradient 2.
\[f(x) = \frac{1}{2}(x^2 + 10x + 25) - 8\]\[f'(x) = x + 5\]\[x + 5 = 2\]\[x = -3\]\[f(-3) = \frac{1}{2}(2)^2 - 8 = -6\]\((-3\,;\,-6)\)
Other points that earn marks
- Alternative: set \(f(x) = 2x - 9 + k\) and use \(\Delta = 0\) for one point of contact.
More Questions On Functions and Inverse Functions
The axes of symmetry pass through the point where the asymptotes meet, \((-2\,;\,-1)\).
With gradient 1:
\[y = x + c\]\[-1 = -2 + c, \text{ so } c = 1\]\(h(x) = x + 1\)
Swap \(x\) and \(y\):
\[x = -6y^2\]\[y^2 = -\frac{x}{6}\]\[y = \pm\sqrt{-\frac{x}{6}}\]Since \(f^{-1}(x) \leq 0\), take the negative root.
\(y = -\sqrt{-\frac{x}{6}}\), \(x \leq 0\)
\(D(-5\,;\,12)\) is on \(f\), and by symmetry about \(x = -3\) the point \((-1\,;\,12)\) is also on \(f\).
\(f\) lies above \(y = 12\) between these points.
\(-5 \lt x \lt -1\)
\(A(-6\,;\,-6)\), \(B(0\,;\,6)\)
\[AB = \sqrt{(0 - (-6))^2 + (6 - (-6))^2}\]\[AB = \sqrt{36 + 144} = \sqrt{180}\]\(AB = 6\sqrt{5} \approx 13{,}42\) units
The axes of symmetry pass through the point where the asymptotes meet, \((1\,;\,2)\), with gradients \(1\) and \(-1\).
\[y - 2 = -(x - 1)\]\(y = -x + 3\)
For the maximum:
\[D'(x) = -2x + 1 = 0 \text{, so } x = \frac{1}{2}\](or \(x = -\frac{b}{2a} = -\frac{1}{2(-1)} = \frac{1}{2}\))
\[D\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 6\]Maximum distance \(= \frac{25}{4} = 6{,}25\) units