أتدرب وأحل المسائل

أتدرب وأحل المسائل

المماس والعمودي على المماس

أجد معادلة المماس لمنحنى كل اقتران ممّا يأتي عند النقطة المعطاة:

(1) $f (x) = x^{3} - 6 x + 3, (2, - 1)$

$\begin{aligned} f (x) = x^{3} - 6 x + 3, (2, - 1), f (2) = - 1 \\ f^{'} (x) = 3 x^{2} - 6 \\ f^{'} (2) = 12 - 6 = 6 \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (2) = f^{'} (2) (x - 2) \\ y - (- 1) = 6 (x - 2) \\ y + 1 = 6 x - 12 \\ y = 6 x - 13 \end{aligned}$

(2) $f (x) = \frac{x^{4} - 3 x^{3}}{x}, (1, - 2)$

$\begin{aligned} f (x) = \frac{x^{4} - 3 x^{3}}{x} = \frac{x^{4}}{x} - \frac{3 x^{3}}{x} = x^{3} - 3 x^{2}, (1, - 2), f (1) = - 2 \\ f^{'} (x) = 3 x^{2} - 6 x \\ f^{'} (1) = 3 - 6 = - 3 \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (1) = f^{'} (1) (x - 1) \\ y - (- 2) = - 3 (x - 1) \\ y + 2 = - 3 x + 3 \\ y = - 3 x + 1 \end{aligned}$

(3) $f (x) = \sqrt{x} (x^{2} - 1), (1, 0)$

$\begin{aligned} f (x) = \sqrt{x} (x^{2} - 1), (1, 0), f (1) = 0 \\ f^{'} (x) = (\sqrt{x}) (2 x) + (x^{2} - 1) (\frac{1}{2 \sqrt{x}}) \\ f^{'} (1) = (1) (2) + (0) (\frac{1}{2}) = 2 \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (1) = f^{'} (1) (x - 1) \\ y - 0 = 2 (x - 1) \\ y = 2 x - 2 \end{aligned}$

(4) $f (x) = x + \frac{4}{x}, (- 4, - 5)$

$\begin{aligned} f (x) = x + \frac{4}{x}, (- 4, - 5), f (- 4) = - 5 \\ f^{'} (x) = 1 - \frac{4}{x^{2}} \\ f^{'} (- 4) = 1 - \frac{4}{16} = 1 - \frac{1}{4} = \frac{3}{4} \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (- 4) = f^{'} (- 4) (x - (- 4)) \\ y - (- 5) = \frac{3}{4} (x + 4) \\ y + 5 = \frac{3}{4} x + 3 \\ y = \frac{3}{4} x - 2 \end{aligned}$

(5) $f (x) = x + e^{x}, (0, 1)$

$\begin{aligned} f (x) = x + e^{x}, (0, 1), f (0) = 1 \\ f^{'} (x) = 1 + e^{x} \\ f^{'} (0) = 1 + e^{0} = 1 + 1 = 2 \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (0) = f^{'} (0) (x - 0) \\ y - 1 = 2 (x - 0) \\ y - 1 = 2 x \\ y = 2 x + 1 \end{aligned}$

(6) $f (x) = \ln (x + e), (0, 1)$

$\begin{aligned} f (x) = l n (x + e), (0, 1), f (0) = 1 \\ f^{'} (x) = \frac{1}{x + e} \\ f^{'} (0) = \frac{1}{0 + e} = \frac{1}{e} \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (0) = f^{'} (0) (x - 0) \\ y - 1 = \frac{1}{e} (x - 0) \\ y - 1 = \frac{1}{e} x \\ y = \frac{1}{e} x + 1 \end{aligned}$

أجد معادلة المماس لمنحنى كل اقتران ممّا يأتي عند قيمة x المعطاة:

(7) $f (x) = \sqrt{x - 7}, x = 16$

$\begin{aligned} f (x) = \sqrt{x - 7}, x = 16 \\ f (16) = \sqrt{16 - 7} = 3 \to (16, 3) \\ f^{'} (x) = \frac{1}{2 \sqrt{x - 7}} \\ f^{'} (16) = \frac{1}{2 \sqrt{16 - 7}} = \frac{1}{6} \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (16) = f^{'} (16) (x - 16) \\ y - 3 = \frac{1}{6} (x - 16) \\ y - 3 = \frac{1}{6} x - \frac{8}{3} \\ y = \frac{1}{6} x + \frac{1}{3} \end{aligned}$

(8) $f (x) = (x - 1) e^{x}, x = 1$

$\begin{aligned} f (x) = (x - 1) e^{x}, x = 1 \\ f (1) = (1 - 1) e^{1} = 0 \to (1, 0) \\ f^{'} (x) = (x - 1) e^{x} + e^{x} (1) = x e^{x} \\ f^{'} (1) = 1 e^{1} = e \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (1) = f^{'} (1) (x - 1) \\ y - 0 = e (x - 1) \\ y = e x - e \end{aligned}$

(9) $f (x) = \frac{x + 3}{x - 3}, x = 4$

$\begin{aligned} f (x) = \frac{x + 3}{x - 3}, x = 4 \\ f (4) = \frac{4 + 3}{4 - 3} = 7 \\ f^{'} (x) = \frac{(x - 3) (1) - (x + 3) (1)}{(x - 3)^{2}} = \frac{- 6}{(x - 3)^{2}} \\ f^{'} (4) = \frac{- 6}{(4 - 3)^{2}} = - 6 \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (4) = f^{'} (4) (x - 4) \\ y - 7 = - 6 (x - 4) \\ y - 7 = - 6 x + 24 \\ y = - 6 x + 31 \end{aligned}$

(10) $f (x) = (\ln x)^{2}, x = e$

$\begin{aligned} f (x) = (l n x)^{2}, x = e \\ f (e) = (l n e)^{2} = 1 \to (e, 1) \\ f^{'} (x) = 2 (l n x) (\frac{1}{x}) \\ f^{'} (e) = 2 (l n e) (\frac{1}{e}) = \frac{2}{e} \end{aligned}$

معادلة المماس:

$\begin{aligned} y - f (a) = f^{'} (a) (x - a) \\ y - f (e) = f^{'} (e) (x - e) \\ y - 1 = \frac{2}{e} (x - e) \\ y - 1 = \frac{2}{e} x + 2 \\ y = \frac{2}{e} x + 3 \end{aligned}$

إعداد : شبكة منهاجي التعليمية

10 / 07 / 2023

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حمّل تطبيق منهاجي الجديد

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