2. Given \( f(x)=\sqrt{3 x+4} \), state the domain and range of \( f(x) \). Hence sketch the graph of \( f(x) \) and \( f^{-1}(x) \) on the same axes.

To find the domain and range of the function \( f(x) = \sqrt{3x + 4} \), we need to consider the constraints imposed by the square root. ### Domain The square root function is defined only for non-negative arguments. Therefore, the expression inside the square root, \( 3x + 4 \), must be greater than or equal to zero: \[ 3x + 4 \geq 0 \] Solving for \( x \): \[ 3x \geq -4 \] \[ x \geq -\frac{4}{3} \] So, the domain of \( f(x) \) is \( x \geq -\frac{4}{3} \), or in interval notation: \[ D_f = \left[-\frac{4}{3}, \infty\right) \] ### Range The square root function always returns non-negative values. Since the minimum value of \( 3x + 4 \) in the domain is 0 (when \( x = -\frac{4}{3} \)), the minimum value of \( f(x) \) is 0. As \( x \) increases, \( 3x + 4 \) increases, and so does \( f(x) \). Therefore, the range of \( f(x) \) is all non-negative real numbers: \[ R_f = [0, \infty) \] ### Graphing \( f(x) \) and \( f^{-1}(x) \) To graph \( f(x) \) and its inverse \( f^{-1}(x) \) on the same axes, we can follow these steps: 1. **Graph \( f(x) \)**: Plot the graph of \( f(x) = \sqrt{3x + 4} \) for \( x \geq -\frac{4}{3} \). The graph will be a curve starting at the point \( \left(-\frac{4}{3}, 0\right) \) and increasing as \( x \) increases. 2. **Reflect \( f(x) \) across the line \( y = x \)**: To graph \( f^{-1}(x) \), we reflect the graph of \( f(x) \) across the line \( y = x \). This will flip the curve over the line \( y = x \), and the resulting graph will be the inverse function. Here's a rough sketch of the graph: ``` y | | * | / | / | / | / | / | / |/______________________ x -4/3 ``` The curve starts at \( \left(-\frac{4}{3}, 0\right) \) and increases to the right, reflecting the domain and range we've determined. Remember, this is a rough sketch, and the actual graph would be smoother and more precise.