أتدرب وأحل المسائل

المماس والعمودي على المماس

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

(1) $f\left(x\right)=x^3-6x+3 , (2,-1)$

$\begin{array}{rl}& f\left(x\right)=x^3-6x+3 , (2,-1) , f\left(2\right)=-1\\ & f^{\mathrm{\prime }}\left(x\right)=3x^2-6\\ & f^{\mathrm{\prime }}\left(2\right)=12-6=6\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(2\right)=f^{\mathrm{\prime }}\left(2\right)(x-2)\\ & y-(-1)=6(x-2)\\ & y+1=6x-12\\ & y=6x-13\end{array}$

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

$\begin{array}{rl}& f\left(x\right)=\frac{x^4-3x^3}{x}=\frac{x^4}{x}-\frac{3x^3}{x}=x^3-3x^2,(1,-2),f\left(1\right)=-2\\ & f^{\mathrm{\prime }}\left(x\right)=3x^2-6x\\ & f^{\mathrm{\prime }}\left(1\right)=3-6=-3\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(1\right)=f^{\mathrm{\prime }}\left(1\right)(x-1)\\ & y-(-2)=-3(x-1)\\ & y+2=-3x+3\\ & y=-3x+1\end{array}$

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

$\begin{array}{rl}& f\left(x\right)=\sqrt{x}(x^2-1),(1,0),f\left(1\right)=0\\ & f^{\mathrm{\prime }}\left(x\right)=\left(\sqrt{x}\right)\left(2x\right)+(x^2-1)\left(\frac{1}{2\sqrt{x}}\right)\\ & f^{\mathrm{\prime }}\left(1\right)=\left(1\right)\left(2\right)+\left(0\right)\left(\frac{1}{2}\right)=2\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(1\right)=f^{\mathrm{\prime }}\left(1\right)(x-1)\\ & y-0=2(x-1)\\ & y=2x-2\end{array}$

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

$\begin{array}{rl}& f\left(x\right)=x+\frac{4}{x},(-4,-5),f(-4)=-5\\ & f^{\mathrm{\prime }}\left(x\right)=1-\frac{4}{x^2}\\ & f^{\mathrm{\prime }}(-4)=1-\frac{4}{16}=1-\frac{1}{4}=\frac{3}{4}\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f(-4)=f^{\mathrm{\prime }}(-4)(x-(-4\left)\right)\\ & y-(-5)=\frac{3}{4}(x+4)\\ & y+5=\frac{3}{4}x+3\\ & y=\frac{3}{4}x-2\end{array}$

(5) $f\left(x\right)=x+e^x , (0,1)$

$\begin{array}{rl}& f\left(x\right)=x+e^x,(0,1) , f\left(0\right)=1\\ & f^{\mathrm{\prime }}\left(x\right)=1+e^x\\ & f^{\mathrm{\prime }}\left(0\right)=1+e^0=1+1=2\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(0\right)=f^{\mathrm{\prime }}\left(0\right)(x-0)\\ & y-1=2(x-0)\\ & y-1=2x\\ & y=2x+1\end{array}$

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

$\begin{array}{rl}& f\left(x\right)=ln(x+e) , (0,1) , f\left(0\right)=1\\ & f^{\mathrm{\prime }}\left(x\right)=\frac{1}{x+e}\\ & f^{\mathrm{\prime }}\left(0\right)=\frac{1}{0+e}=\frac{1}{e}\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(0\right)=f^{\mathrm{\prime }}\left(0\right)(x-0)\\ & y-1=\frac{1}{e}(x-0)\\ & y-1=\frac{1}{e}x\\ & y=\frac{1}{e}x+1\end{array}$

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

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

$\begin{array}{rl}& f\left(x\right)=\sqrt{x-7} , x=16\\ & f\left(16\right)=\sqrt{16-7}=3\to (16,3)\\ & f^{\mathrm{\prime }}\left(x\right)=\frac{1}{2\sqrt{x-7}}\\ & f^{\mathrm{\prime }}\left(16\right)=\frac{1}{2\sqrt{16-7}}=\frac{1}{6}\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(16\right)=f^{\mathrm{\prime }}\left(16\right)(x-16)\\ & y-3=\frac{1}{6}(x-16)\\ & y-3=\frac{1}{6}x-\frac{8}{3}\\ & y=\frac{1}{6}x+\frac{1}{3}\end{array}$

(8) $f\left(x\right)=(x-1)e^x , x=1$

$\begin{array}{rl}& f\left(x\right)=(x-1)e^x , x=1\\ & f\left(1\right)=(1-1)e^1=0\to (1,0)\\ & f^{\mathrm{\prime }}\left(x\right)=(x-1)e^x+e^x\left(1\right)=x{e}^x\\ & f^{\mathrm{\prime }}\left(1\right)=1e^1=e\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(1\right)=f^{\mathrm{\prime }}\left(1\right)(x-1)\\ & y-0=e(x-1)\\ & y=ex-e\end{array}$

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

$\begin{array}{rl}& f\left(x\right)=\frac{x+3}{x-3} , x=4\\ & f\left(4\right)=\frac{4+3}{4-3}=7\\ & f^{\mathrm{\prime }}\left(x\right)=\frac{(x-3)\left(1\right)-(x+3)\left(1\right)}{(x-3{)}^2}=\frac{-6}{(x-3{)}^2}\\ & f^{\mathrm{\prime }}\left(4\right)=\frac{-6}{(4-3{)}^2}=-6\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(4\right)=f^{\mathrm{\prime }}\left(4\right)(x-4)\\ & y-7=-6(x-4)\\ & y-7=-6x+24\\ & y=-6x+31\end{array}$

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

$\begin{array}{rl}& f\left(x\right)=(lnx{)}^2,x=e\\ & f\left(e\right)=(lne{)}^2=1\to (e,1)\\ & f^{\mathrm{\prime }}\left(x\right)=2(lnx)\left(\frac{1}{x}\right)\\ & f^{\mathrm{\prime }}\left(e\right)=2(lne)\left(\frac{1}{e}\right)=\frac{2}{e}\end{array}$

معادلة المماس:

$\begin{array}{rl}& y-f\left(a\right)=f^{\mathrm{\prime }}\left(a\right)(x-a)\\ & y-f\left(e\right)=f^{\mathrm{\prime }}\left(e\right)(x-e)\\ & y-1=\frac{2}{e}(x-e)\\ & y-1=\frac{2}{e}x+2\\ & y=\frac{2}{e}x+3\end{array}$