What is the tangent of f(x)=2xsqrt(5-2x),\at x=-1 ?

The tangent of f(x)=2xsqrt(5-2x),\at x=-1 is y=(16)/(sqrt(7))x-2sqrt(7)+(16)/(sqrt(7))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=2xsqrt(5-2x),\at x=-1

tangent of

f (x) = 2 x 5 - 2 x, at x = - 1

y = \frac{16}{7} x - 27 + \frac{16}{7}

Solution steps

Find the tangent point:

(- 1, - 27)

Find the slope of

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

Compute the slope of

f (x) = 2 x 5 - 2 x

x = - 1 : m = \frac{16}{7}

Find the line with slope m=

\frac{16}{7}

and passing through

(- 1, - 27) : y = \frac{16}{7} x - 27 + \frac{16}{7}

y = \frac{16}{7} x - 27 + \frac{16}{7}

\int r d r

y^{^{'}} = 1 + y^{2}

f (x) = 3 x^{2} - x + 3

\frac{d}{d x} (\frac{x - 17}{x - 17})

\int 4 - x^{2} (16 - 4 x^{2}) d x

What is the tangent of f(x)=2xsqrt(5-2x),\at x=-1 ?
The tangent of f(x)=2xsqrt(5-2x),\at x=-1 is y=(16)/(sqrt(7))x-2sqrt(7)+(16)/(sqrt(7))