For the given function, (a) find the slope of the tangent line to the graph at the given point; (b) find the equation of the tangent line. \( g(x)=3 \sqrt{x} \) at \( x=1 \)

To find the slope of the tangent line to the graph at \( x=1 \), we first need to compute the derivative of \( g(x) = 3\sqrt{x} \). The derivative, using the power rule, is \( g'(x) = \frac{3}{2\sqrt{x}} \). Evaluating this at \( x=1 \), we get \( g'(1) = \frac{3}{2\sqrt{1}} = \frac{3}{2} \). So the slope of the tangent line at \( x=1 \) is \( \frac{3}{2} \). Next, we find the equation of the tangent line. The point on the graph at \( x=1 \) is \( g(1) = 3\sqrt{1} = 3 \). Using the point-slope form \( y - y_1 = m(x - x_1) \), where \( m = \frac{3}{2} \), \( x_1 = 1 \), and \( y_1 = 3 \), we get the equation of the tangent line as \( y - 3 = \frac{3}{2}(x - 1) \). Simplifying this gives \( y = \frac{3}{2}x + \frac{3}{2} \). Thus, the equation of the tangent line is \( y = \frac{3}{2}x + \frac{3}{2} \).