What is the tangent of f(x)=-x^3+x^2+4x-2,\at x=0 ?

The tangent of f(x)=-x^3+x^2+4x-2,\at x=0 is y=4x-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)=-x^3+x^2+4x-2,\at x=0

tangent of

f (x) = - x^{3} + x^{2} + 4 x - 2, at x = 0

y = 4 x - 2

Solution steps

Find the tangent point:

(0, - 2)

Find the slope of

f (x) = - x^{3} + x^{2} + 4 x - 2 : \frac{df ( x )}{d x} = - 3 x^{2} + 2 x + 4

Compute the slope of

f (x) = - x^{3} + x^{2} + 4 x - 2

x = 0 : m = 4

Find the line with slope m=

4

and passing through

(0, - 2) : y = 4 x - 2

y = 4 x - 2

\frac{\partial}{\partial y} (y e^{x y} sin (y))

x \to 2 - lim (3 x)

x \to 0 + lim (x^{3} cot (x))

5 x^{^{''}} + 50 x - 10 = 0

\frac{\partial}{\partial y} ((2 x + 3 y)^{7})

What is the tangent of f(x)=-x^3+x^2+4x-2,\at x=0 ?
The tangent of f(x)=-x^3+x^2+4x-2,\at x=0 is y=4x-2