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6) If the curve of the function \( f: f(x)=x^{2}-1 \) passes through the two points \( \mathrm{A}(2, f(2)) \) , \( \mathrm{B}(3, f(3)) \), then \( \frac{\text { the slope of the secant } \stackrel{\rightharpoonup}{\mathrm{AB}}}{\text { the slope of the tangent at } \mathrm{A}}=\cdots \cdots \cdots \cdots \) \( \begin{array}{llll}\text { (a) } 5 & \text { (b) } \frac{5}{4} & \text { (c) } \frac{4}{5} & \text { (d) } 1\end{array} \)

Ask by Byrd Lowe. in Egypt
Mar 10,2025

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Answer

The ratio of the slope of the secant line \( \stackrel{\rightharpoonup}{\mathrm{AB}} \) to the slope of the tangent at point \( \mathrm{A} \) is \( \frac{5}{4} \).

Solution

1. Compute the coordinates of points \( \mathrm{A} \) and \( \mathrm{B} \): - \( f(x) = x^2 - 1 \) - For \( \mathrm{A}(2, f(2)) \): \[ f(2) = 2^2 - 1 = 4 - 1 = 3 \] So, \( \mathrm{A}(2, 3) \). - For \( \mathrm{B}(3, f(3)) \): \[ f(3) = 3^2 - 1 = 9 - 1 = 8 \] So, \( \mathrm{B}(3, 8) \). 2. Calculate the slope of the secant line \( \stackrel{\rightharpoonup}{\mathrm{AB}} \): \[ m_{\text{sec}} = \frac{f(3) - f(2)}{3 - 2} = \frac{8 - 3}{1} = 5 \] 3. Find the slope of the tangent to the curve at \( \mathrm{A} \): - Differentiate \( f(x) \): \[ f'(x) = 2x \] - Evaluate at \( x = 2 \): \[ m_{\text{tangent}} = f'(2) = 2(2) = 4 \] 4. Compute the ratio of the slopes: \[ \frac{m_{\text{sec}}}{m_{\text{tangent}}} = \frac{5}{4} \] Thus, the answer is \( \frac{5}{4} \).

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To find the slope of the secant line between points A(2, f(2)) and B(3, f(3)), we first compute the function values: \( f(2) = 2^2 - 1 = 3 \) and \( f(3) = 3^2 - 1 = 8 \). The slope of secant line \(\frac{f(3) - f(2)}{3 - 2} = \frac{8 - 3}{1} = 5\). Next, the slope of the tangent line at A can be found by taking the derivative of \( f(x) \): \( f'(x) = 2x \). Evaluating at \( x = 2 \): \( f'(2) = 2(2) = 4 \). Now, we compute the ratio of the slopes: \(\frac{\text{slope of secant}}{\text{slope of tangent}} = \frac{5}{4}\). Thus, the answer is \(\frac{5}{4}\) (b). As a side note, the concept of secant and tangent lines is key in calculus and helps to visualize the transition from average rate of change to the instantaneous rate of change—fascinating, right? Speaking of which, think of how you're constantly using these notions in real life: driving speed (tangent) versus average speed over a trip (secant)!

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