Question

Evaluate the function as indicated. Determine its domain and range. \(f(x)=\left\\{\begin{array}{l}\sqrt{x+4}, x \leq 5 \\ (x-5)^{2}, x>5\end{array}\right.\) (a) \(f(-3)\) (b) \(f(0)\) (c) \(f(5)\) (d) \(f(10)\)

Short answer

(a) \(f(-3) = 1\), (b) \(f(0) = 2\), (c) \(f(5) = 3\), (d) \(f(10) = 25\), domain is \(x \geq -4\), and range is \(y \geq 0\).

Step-by-step solution

Evaluate the Function at Different Values

Evaluate the function at each specified x-value. Use the top equation when \(x \leq 5\) and the bottom equation when \(x>5\). (a) For \(f(-3)\), use the first equation, \(\sqrt{x+4}\) and replace x with -3: \(f(-3) = \sqrt{-3+4} = \sqrt{1} = 1\)(b) For \(f(0)\), use the first equation again: \(f(0) = \sqrt{0+4} = \sqrt{4} = 2\)(c) For \(f(5)\), use the first equation, because \(x\) equals 5: \(f(5) = \sqrt{5+4} = \sqrt{9} = 3\)(d) For \(f(10)\), use the second equation, \((x-5)^{2}\), because \(x\) is more than 5: \(f(10) = (10-5)^{2} = 5^{2} = 25\)

Determine Domain and Range of the Function

The domain of a function consists of all possible x-values. The domain of the first part, \(\sqrt{x+4}\), is \(x \geq -4\). The domain of the second part, \((x-5)^{2}\), is \(x>5\). Combine these two domains, and the overall domain is \(x \geq -4\). The range of a function consists of all possible values of the function. The first part, having a square root, will always be greater than or equal to 0, and the second part, a square, will also always be greater than or equal to 0. Therefore, in both parts the minimum value is 0. As there is no bound on how high either equation can go, the range is \(y \geq 0\).

Summarize Findings

After calculating the function at the given x-values and determining the domain and range, we found: (a) \(f(-3) = 1\), (b) \(f(0) = 2\), (c) \(f(5) = 3\), (d) \(f(10) = 25\), domain is \(x \geq -4\), and range is \(y \geq 0\).