What is the tangent of f(x)= x/(x-3),\at x=0 ?

The tangent of f(x)= x/(x-3),\at x=0 is y=-1/3 x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)= x/(x-3),\at x=0

tangent of

f (x) = \frac{x}{x - 3}, at x = 0

y = - \frac{1}{3} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(0, 0) : y = - \frac{1}{3} x

y = - \frac{1}{3} x

(sin (x))^{3}

f (x) = - 4 x

\int (- sin (x)) d x

\frac{d}{d x} (ln (\frac{x + 1}{x}))

\int_{0}^{6} 6 x - x^{2} d x

What is the tangent of f(x)= x/(x-3),\at x=0 ?
The tangent of f(x)= x/(x-3),\at x=0 is y=-1/3 x