Question

for the following function, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at the indicated point. f(x)=6 cos x at x = π/2. complete the table below. (round the final answer to three decimal places as needed. round all intermediate values to four decimal places as needed.) interval slope of secant line π/2, π/2 + 0.1 -3.820

Explanation:

Step1: セカント線の傾きの公式を用いる

セカント線の傾きは $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$ です。ここで $f(x)=6\cos x$、$x_1=\frac{\pi}{2}$、$x_2=\frac{\pi}{2}+0.1$ です。

Step2: $f(x_1)$ を計算する

$f(x_1)=6\cos(\frac{\pi}{2})=6\times0 = 0$

Step3: $f(x_2)$ を計算する

$f(x_2)=6\cos(\frac{\pi}{2}+0.1)=6\times(-\sin(0.1))\approx6\times(- 0.0998)= - 0.5988$

Step4: セカント線の傾きを計算する

傾き $m=\frac{f(x_2)-f(x_1)}{x_2 - x_1}=\frac{-0.5988 - 0}{(\frac{\pi}{2}+0.1)-\frac{\pi}{2}}=\frac{-0.5988}{0.1}=- 5.988$

Answer:

$-5.988$