What is the tangent of f(x)=sqrt(7x+1),\at x=9 ?

The tangent of f(x)=sqrt(7x+1),\at x=9 is y= 7/16 x+65/16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(7x+1),\at x=9

tangent of

f (x) = 7 x + 1, at x = 9

y = \frac{7}{16} x + \frac{65}{16}

Solution steps

Find the tangent point:

(9, 8)

Find the slope of

f (x) = 7 x + 1 : \frac{df ( x )}{d x} = \frac{7}{2 7 x + 1}

Compute the slope of

f (x) = 7 x + 1

x = 9 : m = \frac{7}{16}

Find the line with slope m=

\frac{7}{16}

and passing through

(9, 8) : y = \frac{7}{16} x + \frac{65}{16}

y = \frac{7}{16} x + \frac{65}{16}

n = 0 \sum \infty \frac{3}{n ( n + 3 )}

\int \frac{2 x + 1}{x ^{6} + 18 x ^{4} + 81 x ^{2}} d x

n = 1 \sum \infty \frac{5}{n ^{2} + 9 n + 7}

x \to \frac{π}{2} lim (x sin (x))

\frac{d}{d x} (e^{2.4})

What is the tangent of f(x)=sqrt(7x+1),\at x=9 ?
The tangent of f(x)=sqrt(7x+1),\at x=9 is y= 7/16 x+65/16