What is the tangent of f(x)=x^2,\at x=0.2 ?

The tangent of f(x)=x^2,\at x=0.2 is y=0.4x-0.04

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^2,\at x=0.2

tangent of

f (x) = x^{2}, at x = 0.2

y = 0.4 x - 0.04

Solution steps

Find the tangent point:

(0.2, 0.04)

Find the slope of

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

Compute the slope of

f (x) = x^{2}

x = 0.2 : m = 0.4

Find the line with slope m=

0.4

and passing through

(0.2, 0.04) : y = 0.4 x - 0.04

y = 0.4 x - 0.04

f (x) = \frac{7}{x ^{2}}

\int \frac{1}{3} sec (x) tan (x) d x

\frac{d}{d x} (- \frac{1}{x ^{2}} + 1)

x \to \infty lim (1 - 1^{x})

y = (3 x^{2} - 4 x) e^{5 x}

What is the tangent of f(x)=x^2,\at x=0.2 ?
The tangent of f(x)=x^2,\at x=0.2 is y=0.4x-0.04