1. Problem: Find the slope of the tangent line and the equation of the tangent line for the function $f(x) = x + \frac{2}{x}$ at the point $(1,3)$. 2. Formula: The slope of the tangent line is given by the derivative $f'(x)$ evaluated at the point's $x$-coordinate. 3. Differentiate $f(x)$: $$f'(x) = \frac{d}{dx}\left(x + \frac{2}{x}\right) = 1 - \frac{2}{x^2}$$ 4. Evaluate the slope at $x=1$: $$f'(1) = 1 - \frac{2}{1^2} = 1 - 2 = -1$$ 5. Equation of tangent line using point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 3 = -1(x - 1)$$ $$y - 3 = -x + 1$$ $$y = -x + 4$$ --- 1. Problem: Find the slope and tangent line for $f(x) = x^2 + 3x + 1$ at $(1,5)$. 2. Derivative: $$f'(x) = 2x + 3$$ 3. Evaluate at $x=1$: $$f'(1) = 2(1) + 3 = 5$$ 4. Tangent line: $$y - 5 = 5(x - 1)$$ $$y - 5 = 5x - 5$$ $$y = 5x$$ --- 1. Problem: Find slope and tangent line for $f(x) = -2x^3$ at $(3,-54)$. 2. Derivative: $$f'(x) = -6x^2$$ 3. Evaluate at $x=3$: $$f'(3) = -6(3)^2 = -6 \times 9 = -54$$ 4. Tangent line: $$y + 54 = -54(x - 3)$$ $$y + 54 = -54x + 162$$ $$y = -54x + 108$$ --- 1. Problem: Find slope and tangent line for $f(x) = 3x^2 + 1$ at $(-2,13)$. 2. Derivative: $$f'(x) = 6x$$ 3. Evaluate at $x=-2$: $$f'(-2) = 6(-2) = -12$$ 4. Tangent line: $$y - 13 = -12(x + 2)$$ $$y - 13 = -12x - 24$$ $$y = -12x - 11$$