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}}

Compute the slope of

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

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 9

y = (x - 4)^{2}, at x = 5

\frac{d y}{d t} = - 2 y + 10

\int x^{3} (x^{4} - 1)^{5} d x

y = 4 e^{x} cos (x), (0, 4)

\frac{\partial}{\partial x} (x \cdot e^{y} + ln (x) \cdot y)

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