Find the slope of the graph of the function \( f(x)=\sqrt{3 x} \) at \( (3,3) \) Then find an equation of the line tangent to the graph al that point.

To find the slope of the graph of the function \( f(x) = \sqrt{3x} \) at the point \( (3, 3) \), we first need to calculate the derivative of \( f(x) \). The derivative \( f'(x) \) is given by: \[ f'(x) = \frac{d}{dx}(\sqrt{3x}) = \frac{1}{2\sqrt{3x}} \cdot 3 = \frac{3}{2\sqrt{3x}}. \] Now, substituting \( x = 3 \) into the derivative to find the slope: \[ f'(3) = \frac{3}{2\sqrt{3 \cdot 3}} = \frac{3}{2 \cdot 3} = \frac{1}{2}. \] Thus, the slope at the point \( (3, 3) \) is \( \frac{1}{2} \). Next, we use the point-slope form of the equation of a line to find the tangent line: \[ y - y_1 = m(x - x_1), \] where \( m \) is the slope and \( (x_1, y_1) \) is the point \( (3, 3) \). Plugging in the values: \[ y - 3 = \frac{1}{2}(x - 3). \] Simplifying gives: \[ y - 3 = \frac{1}{2}x - \frac{3}{2} \implies y = \frac{1}{2}x + \frac{3}{2}. \] Therefore, the equation of the tangent line at the point \( (3, 3) \) is: \[ y = \frac{1}{2}x + \frac{3}{2}. \] This means at \( (3, 3) \), the graph of \( f \) is rising, showcasing how determined the square root function is to grow wider and taller! Isn't it wild how a small change in input can lead to such big consequences in output? If you’re interested in understanding more complex derivatives and their applications, consider diving into calculus textbooks or online resources. You’ll uncover the rich tapestry of mathematics not only in graphs but across various fields such as physics and engineering, where rates of change and slopes can describe everything from velocity to optimization problems!