Answer:
[tex]f^{-1}(x)=\dfrac{ \ln \left( \dfrac{x+3}{15} \right) }{ \ln 5}, \quad \textsf{for}\; \{x : x > -3 \}[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=3 \cdot 5^{x+1}-3[/tex]
The domain of the given function is unrestricted: (-∞, +∞).
The range of the given function is restricted: (-3, +∞).
The inverse of a function is its reflection in the line y = x.
To find the inverse of a function, replace x with y:
[tex]\implies x=3 \cdot 5^{y+1}-3[/tex]
Rearrange the equation to make y the subject:
[tex]\implies x=3 \cdot 5^{y+1}-3[/tex]
[tex]\implies x=3 \cdot 5^y \cdot 5^1-3[/tex]
[tex]\implies x=15 \cdot 5^y -3[/tex]
[tex]\implies x+3=15 \cdot 5^y[/tex]
[tex]\implies \dfrac{x+3}{15}= 5^y[/tex]
[tex]\implies \ln \left( \dfrac{x+3}{15} \right) = \ln 5^y[/tex]
[tex]\implies \ln \left( \dfrac{x+3}{15} \right) = y \ln 5[/tex]
[tex]\implies y=\dfrac{ \ln \left( \dfrac{x+3}{15} \right) }{ \ln 5}[/tex]
Replace y with f⁻¹(x):
[tex]\implies f^{-1}(x)=\dfrac{ \ln \left( \dfrac{x+3}{15} \right) }{ \ln 5}[/tex]
The domain of the inverse of a function is the same as the range of the original function.
Therefore, the domain of the inverse function is (-3, +∞).
Therefore, the inverse of the given function is:
[tex]f^{-1}(x)=\dfrac{ \ln \left( \dfrac{x+3}{15} \right) }{ \ln 5}, \quad \textsf{for}\; \{x : x > -3 \}[/tex]
