What is the tangent of f(x)=(20x)/(x^2-5),\at x=5 ?

The tangent of f(x)=(20x)/(x^2-5),\at x=5 is y=-3/2 x+25/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=(20x)/(x^2-5),\at x=5

tangent of

f (x) = \frac{20 x}{x ^{2} - 5}, at x = 5

y = - \frac{3}{2} x + \frac{25}{2}

Solution steps

Find the tangent point:

(5, 5)

Find the slope of

f (x) = \frac{20 x}{x ^{2} - 5} : \frac{df ( x )}{d x} = \frac{20 ( - x ^{2} - 5 )}{( x ^{2} - 5 ) ^{2}}

Compute the slope of

f (x) = \frac{20 x}{x ^{2} - 5}

x = 5 : m = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(5, 5) : y = - \frac{3}{2} x + \frac{25}{2}

y = - \frac{3}{2} x + \frac{25}{2}

\int \frac{1}{2} sec^{2} (x) + 4 x d x

\frac{d ^{3}}{d x ^{3}} (2 x + 8)

\int_{- 5}^{0} (25 - x^{2}) d x

y^{^{'}} = e^{- x}

\frac{\partial}{\partial z} (x y + z^{3})

What is the tangent of f(x)=(20x)/(x^2-5),\at x=5 ?
The tangent of f(x)=(20x)/(x^2-5),\at x=5 is y=-3/2 x+25/2