15-18 A function $y = f(x)$ and an x-value $x_0$ are given. a. Find a formula for the slope of the tangent line to the graph of $f$ at a general point $x = x_0$. b. Use the formula obtained in part (a) to find the slope of the tangent line for the given value of $x_0$. 15. $f(x) = x^2 - 1$; $x_0 = -1$ ? Answer a. $2x_0$ b. -2 ? Solution a. $m_{tan} = \lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)}{x_1 - x_0} = \lim_{x_1 \to x_0} \frac{(x_1^2 - 1) - (x_0^2 - 1)}{x_1 - x_0} = \lim_{x_1 \to x_0} \frac{(x_1^2 - x_0^2)}{x_1 - x_0} = \lim_{x_1 \to x_0} (x_1 + x_0) = 2x_0$ b. $m_{tan} = 2(-1) = -2$ 16. $f(x) = x^2 + 3x + 2$; $x_0 = 2$
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Step 1: Given the function f(x) = x^2 - 1 and x_0 = -1, we need to find the slope of the tangent line at x = x_0. Show more…
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