What is the equation of the line normal to $f (x) = \frac{x}{x^{2} - 2}$ $f(x)=x/(x^2-2)$ at $x = 1$ $x=1$ ?

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What is the equation of the line normal to $f (x) = \frac{x}{x^{2} - 2}$ $f(x)=x/(x^2-2)$ at $x = 1$ $x=1$ ?

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Answer 1 · 2017-11-22T12:11:20

$y = \frac{1}{3} x - \frac{4}{3}$ $y=1/3x-4/3$

Explanation:

In order to find the normal to the tangent we first need to find the gradient of the tangent line at $x = 1$ $x=1$

Differentiate $f (x)$ $f(x)$

$\frac{d}{d x} (x {(x^{2} - 2)}^{- 1}) = {(x^{2} - 2)}^{- 1} - {(x^{2} - 2)}^{- 2} \cdot 2 x \cdot x$ $d/dx(x(x^2-2)^(-1))=(x^2-2)^(-1)-(x^2-2)^(-2)*2x*x$

$\to = \frac{1}{(x^{2} - 2)} - \frac{2 x^{2}}{{(x^{2} - 2)}^{2}} = \frac{(x^{2} - 2) - 2 x^{2}}{{(x^{2} - 2)}^{2}} = \frac{- x^{2} - 2}{{(x^{2} - 2)}^{2}} = - \frac{x^{2} + 2}{{(x^{2} - 2)}^{2}}$ $->=1/((x^2-2))- (2x^2)/(x^2-2)^2=((x^2-2)-2x^2)/(x^2-2)^2=(-x^2-2)/(x^2-2)^2=- (x^2+2)/(x^2-2)^2$

Gradient of tangent at $x = 1$ $x=1$

$- \frac{1^{2} + 2}{{(1^{2} - 2)}^{2}} = - 3$ $- (1^2+2)/(1^2-2)^2=-3$

gradient of normal:

$- 3 \cdot m_{2} = - 1 \Rightarrow m_{2} = \frac{1}{3}$ $-3*m_2=-1=>m_2=1/3$

Coordinates at $x = 1$ $x=1$

$(1, - 1)$ $( 1 , -1 )$

Equation of normal:

$y + 1 = \frac{1}{3} (x - 1) \Rightarrow y = \frac{1}{3} x - \frac{4}{3}$ $y+1=1/3(x-1)=>y=1/3x-4/3$

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What is the equation of the line normal to $f (x) = \frac{x}{x^{2} - 2}$ $f(x)=x/(x^2-2)$ at $x = 1$ $x=1$ ?

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What is the equation of the line normal to f(x)=xx2−2 f(x)=x/(x^2-2) at x=1 x=1?

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What is the equation of the line normal to $f (x) = \frac{x}{x^{2} - 2}$ $f(x)=x/(x^2-2)$ at $x = 1$ $x=1$ ?