91Ó°ÊÓ

The domain of \(g(x) = \frac{1}{2}x - \sqrt{x}\) consists of all the real numbers for which the function is defined. Since we cannot take the square root of a negative number, the domain of the function is \(x \ge 0\).

To find the x-intercept, we set \(g(x) = 0\): \[ \frac{1}{2}x - \sqrt{x} = 0 \] Solving for x, we can factor out a \(\sqrt{x}\): \[ \sqrt{x}(\frac{1}{2}\sqrt{x} - 1) = 0 \] This gives us two possible solutions, \(x = 0\) and \(x = 4\). So, the x-intercepts are \(x = 0\) and \(x = 4\). To find the y-intercept, we set \(x = 0\): \[ g(0) = \frac{1}{2}(0) - \sqrt{0} = 0 \] Thus, the y-intercept is at \(y = 0\).

First, we'll find the derivative of the function \(g(x)\) with respect to \(x\). This will help us determine the critical points of the function: \[ g'(x) = \frac{1}{2} - \frac{1}{2\sqrt{x}} \] Now, we'll set \(g'(x) = 0\) to find the critical points: \[ \frac{1}{2} - \frac{1}{2\sqrt{x}} = 0 \] Solving for x, we find that \(x = 1\). This is our only critical point. We can test intervals around the critical point to determine the intervals of increase and decrease: - For \(x \in (0,1)\), \(g'(x) > 0\), so the function is increasing. - For \(x \in (1,+\infty)\), \(g'(x) < 0\), so the function is decreasing.

To find the concavity and inflection points, we'll find the second derivative of the function \(g(x)\): \[ g''(x) = \frac{1}{4x^{3/2}} \] Since the second derivative is always positive for \(x>0\), the function is concave up for all values of \(x\) in its domain. Now, we'll put all the information together to sketch the graph of the function. The graph will have the following properties: - The function is defined for \(x \ge 0\). - The function has x-intercepts at \(x = 0\) and \(x = 4\); y-intercept at \(y = 0\). - The function is increasing for \(x \in (0,1)\) and decreasing for \(x \in (1,+\infty)\). - The function is concave up for all values of \(x > 0\). Using all this information, sketch the graph of the function \(g(x) = \frac{1}{2}x - \sqrt{x}\).