What is the tangent of f(x)= 1/(2+3x),(2, 1/8) ?

The tangent of f(x)= 1/(2+3x),(2, 1/8) is y=-3/64 x+7/32

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)= 1/(2+3x),(2, 1/8)

tangent of

f (x) = \frac{1}{2 + 3 x}, (2, \frac{1}{8})

y = - \frac{3}{64} x + \frac{7}{32}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \frac{1}{2 + 3 x}

(2, \frac{1}{8}) : m = - \frac{3}{64}

Find the line with slope m=

- \frac{3}{64}

and passing through

(2, \frac{1}{8}) : y = - \frac{3}{64} x + \frac{7}{32}

y = - \frac{3}{64} x + \frac{7}{32}

y = 1 - 2 x

g (x) = (x^{2} + 1)^{3} (x^{2} + 2)^{6}

\frac{s + 3}{s ^{2} + 2 s + 1}

f (x) = 4 x^{4} - \frac{1}{4 x ^{4}}

y^{^{''}} + 2 y^{^{'}} + y = - e^{- t}

What is the tangent of f(x)= 1/(2+3x),(2, 1/8) ?
The tangent of f(x)= 1/(2+3x),(2, 1/8) is y=-3/64 x+7/32