What is the tangent of f(x)=(20)/(sqrt(x-1)),\at x=5 ?

The tangent of f(x)=(20)/(sqrt(x-1)),\at x=5 is y=-5/4 x+65/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=(20)/(sqrt(x-1)),\at x=5

tangent of

f (x) = \frac{20}{x - 1}, at x = 5

y = - \frac{5}{4} x + \frac{65}{4}

Solution steps

Find the tangent point:

(5, 10)

Find the slope of

f (x) = \frac{20}{x - 1} : \frac{df ( x )}{d x} = - \frac{10}{( x - 1 ) ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{20}{x - 1}

x = 5 : m = - \frac{5}{4}

Find the line with slope m=

- \frac{5}{4}

and passing through

(5, 10) : y = - \frac{5}{4} x + \frac{65}{4}

y = - \frac{5}{4} x + \frac{65}{4}

t^{2} t^{3}

\int_{0}^{\infty} x e^{- 2 x} d x

\int (- 4 x y + 2 x^{3} - 2 y + 3) d x

\int (sin (wt))^{2} d t

\int (2^{x} + e^{x}) d x

What is the tangent of f(x)=(20)/(sqrt(x-1)),\at x=5 ?
The tangent of f(x)=(20)/(sqrt(x-1)),\at x=5 is y=-5/4 x+65/4