Complete the following for the piecewise-defined function \( f(x)=\left\{\begin{array}{rll}x^{2} & \text { if } & x<0 \\ 4 & \text { if } & x=0 \\ \sqrt{x} & \text { if } & x>0\end{array}\right. \) a. Find \( f(-4), f(0) \), and \( f(10) \). b. Sketch the graph of \( y=f(x) \). c. Find the domain of \( f \). d. Find the range of \( f \).

Let's analyze the piecewise-defined function \( f(x) \) step by step. ### a. Find \( f(-4), f(0) \), and \( f(10) \). 1. **Finding \( f(-4) \)**: Since \( -4 < 0 \), we use the first piece of the function: \[ f(-4) = (-4)^2 = 16. \] 2. **Finding \( f(0) \)**: Since \( x = 0 \), we use the second piece of the function: \[ f(0) = 4. \] 3. **Finding \( f(10) \)**: Since \( 10 > 0 \), we use the third piece of the function: \[ f(10) = \sqrt{10}. \] Thus, we have: - \( f(-4) = 16 \) - \( f(0) = 4 \) - \( f(10) = \sqrt{10} \) ### b. Sketch the graph of \( y=f(x) \). To sketch the graph of \( f(x) \): - For \( x < 0 \): The function is \( f(x) = x^2 \). This is a parabola opening upwards, starting from the point \( (0, 4) \) and going to the left. - At \( x = 0 \): The function has a value of \( f(0) = 4 \). This is a single point on the graph. - For \( x > 0 \): The function is \( f(x) = \sqrt{x} \). This is the right half of the square root function, starting from the point \( (0, 0) \) and increasing. The graph will look like this: - A parabola for \( x < 0 \) that approaches \( (0, 4) \) as \( x \) approaches 0 from the left. - A point at \( (0, 4) \). - A curve starting from \( (0, 0) \) and increasing for \( x > 0 \). ### c. Find the domain of \( f \). The domain of \( f \) is the set of all \( x \) values for which \( f(x) \) is defined. - For \( x < 0 \), \( f(x) = x^2 \) is defined for all negative \( x \). - For \( x = 0 \), \( f(x) = 4 \) is defined. - For \( x > 0 \), \( f(x) = \sqrt{x} \) is defined for all positive \( x \). Thus, the domain of \( f \) is: \[ \text{Domain} = (-\infty, 0] \cup (0, \infty). \] ### d. Find the range of \( f \). To find the range of \( f \): 1. For \( x < 0 \): \( f(x) = x^2 \) takes values from \( 0 \) to \( 16 \) (as \( x \) approaches \( 0 \) from the left, \( f(x) \) approaches \( 4 \)). 2. At \( x = 0 \): \( f(0) = 4 \). 3. For \( x > 0 \): \( f(x) = \sqrt{x} \) takes values from \( 0 \) to \( \infty \) (as \( x \) increases). Combining these, the minimum value of \( f(x) \) is \( 0 \) (from the square root function) and the maximum value is unbounded as \( x \) increases. The value \( 4 \) is included in the range. Thus, the range of \( f \) is: \[ \text{Range} = [0, \infty). \] ### Summary of Results: - \( f(-4) = 16 \) - \( f(0) = 4 \) - \( f(10) = \sqrt{10} \) - Domain: \( (-\infty, 0] \cup (0, \infty) \) - Range: \( [0, \infty) \)