What is the tangent of f(x)=x^3+x^2+6/x ,\at x=1 ?

The tangent of f(x)=x^3+x^2+6/x ,\at x=1 is y=-x+9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^3+x^2+6/x ,\at x=1

tangent of

f (x) = x^{3} + x^{2} + \frac{6}{x}, at x = 1

y = - x + 9

Solution steps

Find the tangent point:

(1, 8)

Find the slope of

f (x) = x^{3} + x^{2} + \frac{6}{x} : \frac{df ( x )}{d x} = 3 x^{2} + 2 x - \frac{6}{x ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 8) : y = - x + 9

y = - x + 9

(- 4, 6), (10, - 1)

y^{^{'}} = \frac{( x ^{2} e ^{\frac{y}{x}} + y ^{2} )}{x y}

\frac{\partial}{\partial x} (2 x y + sin (x))

\frac{( e ^{- s} )}{s ( s + 1 )}

\frac{d}{d x} (128 sin (2 x))

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