What is the normal of f(x)=(27)/(5x^2)-3/2 ,\at x=1 ?

The normal of f(x)=(27)/(5x^2)-3/2 ,\at x=1 is f(x)= 5/54 x+514/135

Popular >

normal of f(x)=(27)/(5x^2)-3/2 ,\at x=1

normal of

f (x) = \frac{27}{5 x ^{2}} - \frac{3}{2}, at x = 1

f (x) = \frac{5}{54} x + \frac{514}{135}

Solution steps

Find the tangent point:

(1, \frac{39}{10})

Find the slope of

f (x) = \frac{27}{5 x ^{2}} - \frac{3}{2} : \frac{df ( x )}{d x} = - \frac{54}{5 x ^{3}}

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 = - \frac{54}{5}

Compute the slope of the perpendicular line:

m_{p} = \frac{5}{54}

Find the line with slope m=

\frac{5}{54}

and passing through

(1, \frac{39}{10}) : f (x) = \frac{5}{54} x + \frac{514}{135}

f (x) = \frac{5}{54} x + \frac{514}{135}

\int_{1}^{5} \frac{x ^{2} + 5}{6 x - x ^{2}} d x

\frac{d y}{d x} = \frac{( x y + x )}{y}

y^{^{''}} + 2 π y^{^{'}} + π^{2} y = 0

\frac{d y}{d x} = \frac{x ^{2} - 2}{3 y}

\int \frac{( x ^{4} + 2 )}{x ^{3} + 9 x} d x

What is the normal of f(x)=(27)/(5x^2)-3/2 ,\at x=1 ?
The normal of f(x)=(27)/(5x^2)-3/2 ,\at x=1 is f(x)= 5/54 x+514/135