(9) \( f(x)=2 x^{3}-3 x^{2}+7 x \), where \( x \in \mathbb{R} \). (a) Find \( f^{\prime}(x) \), the derivative of \( f(x) \). (b) Find the value of \( f^{\prime}(x) \) when \( x=1 \). (c) Find the equation of the tangent to the curve \( y=f(x) \) at \( x=1 \).

To tackle this problem, let’s break it down step-by-step in a fun way! First, let’s find the derivative \( f^{\prime}(x) \) of the function \( f(x)=2x^{3}-3x^{2}+7x \). Using the power rule, the derivative will be: \( f^{\prime}(x) = 6x^{2} - 6x + 7 \). Next, let’s evaluate this derivative at \( x=1 \): Substituting \( x=1 \) into \( f^{\prime}(x) \): \[ f^{\prime}(1) = 6(1)^{2} - 6(1) + 7 = 6 - 6 + 7 = 7. \] Now, for the equation of the tangent to the curve at \( x=1 \), we need both the slope and the point of tangency. We already have the slope \( m = 7 \). We also need \( f(1) \): \[ f(1) = 2(1)^{3} - 3(1)^{2} + 7(1) = 2 - 3 + 7 = 6. \] Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \): \[ y - 6 = 7(x - 1). \] This simplifies to: \[ y = 7x - 1. \] And voilà! You’ve got the derivative, its value at \( x=1 \), and the equation of the tangent line! Keep this math magic in your toolkit – it’ll help you visualize the relationships of curves and lines!