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

The tangent of f(x)=1-x^2 is y=-2a_{0}x+1+a_{0}^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=1-x^2

tangent of

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

y = - 2 a_{0} x + 1 + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 1 - a_{0}^{2})

Find the slope of

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

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 = - 2 a_{0}

Find the line with slope m=

- 2 a_{0}

and passing through

(a_{0}, 1 - a_{0}^{2}) : y = - 2 a_{0} x + 1 + a_{0}^{2}

y = - 2 a_{0} x + 1 + a_{0}^{2}

What is the tangent of f(x)=1-x^2 ?
The tangent of f(x)=1-x^2 is y=-2a_{0}x+1+a_{0}^2