What is the normal of f(x)= x/(2x-3),(1,-1) ?

The normal of f(x)= x/(2x-3),(1,-1) is f(x)= 1/3 x-4/3

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normal of f(x)= x/(2x-3),(1,-1)

normal of

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

f (x) = \frac{1}{3} x - \frac{4}{3}

Solution steps

Find the slope of

f (x) = \frac{x}{2 x - 3} : \frac{df ( x )}{d x} = - \frac{3}{( 2 x - 3 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 3

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(1, - 1) : f (x) = \frac{1}{3} x - \frac{4}{3}

f (x) = \frac{1}{3} x - \frac{4}{3}

\int \frac{e ^{- s t}}{s ^{2}} d t

y = \frac{100}{x ^{4}}

f (x) = 14400 + 600 x + x^{2}

\frac{\partial}{\partial x} (sin (2 x) cos (y))

\frac{d y}{d x} = e^{6 x} + 8 y

What is the normal of f(x)= x/(2x-3),(1,-1) ?
The normal of f(x)= x/(2x-3),(1,-1) is f(x)= 1/3 x-4/3