What is the tangent of f(x)=x-x^2,\at x=-1 ?

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

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-x^2,\at x=-1

tangent of

f (x) = x - x^{2}, at x = - 1

y = 3 x + 1

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

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

Compute the slope of

f (x) = x - x^{2}

x = - 1 : m = 3

Find the line with slope m=

3

and passing through

(- 1, - 2) : y = 3 x + 1

y = 3 x + 1

f (x) = 3 x^{5}

\int 1 + x^{2} 2 x d x

\frac{1}{( s + 20 ) [ ( s + 2 ) ^{2} + 1 ]}

x \to \frac{π}{2} lim (x sec^{2} (x))

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