What is the tangent of f(x)=sqrt(x),(9,3) ?

The tangent of f(x)=sqrt(x),(9,3) is y= 1/6 x+3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(x),(9,3)

tangent of

f (x) = x, (9, 3)

y = \frac{1}{6} x + \frac{3}{2}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x

(9, 3) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(9, 3) : y = \frac{1}{6} x + \frac{3}{2}

y = \frac{1}{6} x + \frac{3}{2}

y = \frac{3}{x}, y = 12 x, y = \frac{1}{12} x

y = x (x + 2)

f (x) = \frac{x}{4 + x}

y = ln (8 + ln (x))

\int \frac{1}{x ^{2} - 196} d x

What is the tangent of f(x)=sqrt(x),(9,3) ?
The tangent of f(x)=sqrt(x),(9,3) is y= 1/6 x+3/2