Determine the equation of the inverse of \(h\), in the form \(y = \ldots\)
About this question
Question 4.3.2 of Mathematics Grade 12 Paper 1, NSC Examination (May/June) (DBE, May/June 2022). It tests the Term 1 ATP topic Exponential and Logarithmic Functions.
- TermTerm 1
- ATP topicExponential and Logarithmic Functions
- PaperPaper 1
- Marks2 marks for the method (formula and substitution) and the correct final answer
The graph of \(g(x) = a\left(\frac{1}{3}\right)^x + 7\) passes through the point \(E(-2\,;\,10)\).
For 4.3: \(h(x) = \left(\frac{1}{3}\right)^x\)
Swap \(x\) and \(y\):
\[x = \left(\frac{1}{3}\right)^y\]\(y = \log_{\frac{1}{3}} x\)
Other points that earn marks
- Also accepted: \(y = -\log_{3} x\).
More Questions On Exponential and Logarithmic Functions
Swap \(x\) and \(y\):
\[x = 3^y\]\(g^{-1}: y = \log_{3} x\)
R is the \(y\)-intercept of \(g\): \(y = 2^0 = 1\), so \(R(0\,;\,1)\).
\(TR \parallel x\)-axis, so T has \(y = 1\):
\[1 = \log_2 x\]\[x = 2 \text{, so } T(2\,;\,1)\]\[RT = 2 - 0 = 2 \text{ units}\]\(P'(2\,;\,4)\) is directly above T, so \(P'T = 4 - 1 = 3\) units and \(P'T \perp RT\).
\[\text{Area } \triangle RTP' = \frac{1}{2}(RT)(P'T) = \frac{1}{2}(2)(3)\]Area \(= 3\) square units
Swap \(x\) and \(y\):
\[x = \log_{\frac{1}{3}} y\]\(y = \left(\frac{1}{3}\right)^x\) (or \(y = 3^{-x}\))
Reflection in \(y = x\) swaps the coordinates: \((x\,;\,y) \to (y\,;\,x)\).
\(P'(2\,;\,4)\)
The asymptote is \(y = -16\), so \(q = -16\).
\[f(x) = 2^{x + p} - 16\]Substitute \(T(3\,;\,16)\):
\[16 = 2^{3 + p} - 16\]\[2^{3 + p} = 32 = 2^5\]\[3 + p = 5\]\(p = 2\) and \(q = -16\)