Read an exponential transformation and obtain its logarithmic inverse.
Notes and key methods
For \(\displaystyle f(x)=D+A\cdot 2^{B(x-h)}\) with A,B>0, the domain is all real x, the range is (D,∞), and the horizontal asymptote is \(\displaystyle y=D\). The function increases and never attains its asymptote.
To invert it, subtract D, divide by A, take a base-two logarithm and solve for the original input. The inverse graph reflects the original in \(\displaystyle y=x\), exchanging domain and range.
Check the conditions
The exponential factor is strictly positive.
The inverse logarithm argument must be positive.
Three worked examples
Example 1 · Exponential graphs and inverses
Let \(f(x)=4 \cdot 2^{3 x - 9} + 3\). For short checks use \(\displaystyle h=3, B=3, A=4, D=3\).
Use the labelled scales alongside the numerical method.
State the domain, range and asymptote of f.
Find \(\displaystyle f^{−1}\) and state its domain and range.
Find \(\displaystyle f(h+ \frac{1}{B})\).
Find the x-value where \(\displaystyle f(x)=D+4A\).
Show the reasoning
(a)
Every real input gives a positive exponential term. Thus the output is strictly above D, approaching but never reaching that height.