Logarithmic Inverse 57154A

1. **Problem Statement:** We are given the function $$f(x)=\ln(2x+1)$$ and asked to find: (a) The domain and range of $$f(x)$$. (b) The inverse function $$f^{-1}(x)$$, its domain and range, and the value of $$x$$ such that $$f^{-1}(x)=0$$. (c) Sketch the graphs of $$f(x)$$ and $$f^{-1}(x)$$ on the same axes. --- 2. **Part (a): Domain and Range of $$f(x)$$** - The function $$f(x)=\ln(2x+1)$$ is defined only when the argument of the logarithm is positive: $$2x+1>0$$ $$2x> -1$$ $$x> -\frac{1}{2}$$ - Therefore, the **domain** is: $$\boxed{\left(-\frac{1}{2}, \infty\right)}$$ - The natural logarithm function $$\ln(y)$$ has range $$(-\infty, \infty)$$ for $$y>0$$. - Since $$2x+1$$ can take any positive value from 0 to infinity, the **range** of $$f(x)$$ is: $$\boxed{(-\infty, \infty)}$$ --- 3. **Part (b): Find the inverse function $$f^{-1}(x)$$ and its domain and range** - Start with: $$y = \ln(2x+1)$$ - To find the inverse, swap $$x$$ and $$y$$: $$x = \ln(2y+1)$$ - Exponentiate both sides to remove the logarithm: $$e^x = 2y + 1$$ - Solve for $$y$$: $$2y = e^x - 1$$ $$y = \frac{e^x - 1}{2}$$ - So the inverse function is: $$f^{-1}(x) = \frac{e^x - 1}{2}$$ - **Domain of $$f^{-1}(x)$$:** This is the range of $$f(x)$$, which is: $$(-\infty, \infty)$$ - **Range of $$f^{-1}(x)$$:** This is the domain of $$f(x)$$, which is: $$\left(-\frac{1}{2}, \infty\right)$$ - **Find $$x$$ such that $$f^{-1}(x) = 0$$:** $$0 = \frac{e^x - 1}{2}$$ Multiply both sides by 2: $$0 = e^x - 1$$ $$e^x = 1$$ Taking natural log: $$x = \ln(1) = 0$$ - So, $$\boxed{x=0}$$ satisfies $$f^{-1}(x)=0$$. --- 4. **Part (c): Sketch the graphs of $$f(x)$$ and $$f^{-1}(x)$$** - The graph of $$f(x) = \ln(2x+1)$$ is a logarithmic curve starting from $$x = -\frac{1}{2}$$ and increasing without bound. - The graph of $$f^{-1}(x) = \frac{e^x - 1}{2}$$ is an exponential curve shifted and scaled. - Both graphs are reflections of each other about the line $$y = x$$. - They intersect at the point where $$f(x) = f^{-1}(x)$$, which is at $$x=0, y=0$$. --- **Final answers:** - (a) Domain: $$\left(-\frac{1}{2}, \infty\right)$$, Range: $$(-\infty, \infty)$$ - (b) $$f^{-1}(x) = \frac{e^x - 1}{2}$$, Domain: $$(-\infty, \infty)$$, Range: $$\left(-\frac{1}{2}, \infty\right)$$, and $$f^{-1}(0) = 0$$ at $$x=0$$. - (c) Graphs are reflections about $$y=x$$, both increasing, with the logarithmic and exponential shapes described.