What is the tangent of f(x)= 1/(1+x^2),(2, 2/5) ?

The tangent of f(x)= 1/(1+x^2),(2, 2/5) is y= 3/50 x+7/25 \lor y=-0.31253…x+1.02506…

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)= 1/(1+x^2),(2, 2/5)

tangent of

f (x) = \frac{1}{1 + x ^{2}}, (2, \frac{2}{5})

y = \frac{3}{50} x + \frac{7}{25} or y = - 0.31253 \dots x + 1.02506 \dots

Solution steps

Slope of the tangent line at

(2, \frac{2}{5}) : m = \frac{- 2 x ^{2} + 3}{5 ( x ^{2} + 1 ) ( x - 2 )}

Find the slope of

f (x) = \frac{1}{1 + x ^{2}} : m = - \frac{2 x}{( 1 + x ^{2} ) ^{2}}

Solve

- \frac{2 x}{( 1 + x ^{2} ) ^{2}} = \frac{- 2 x ^{2} + 3}{5 ( x ^{2} + 1 ) ( x - 2 )} : x = - 3, x \approx 0.16487 \dots

Find the tangent line at

x = - 3 : y = \frac{3}{50} x + \frac{7}{25}

Find the tangent line at

x = 0.16487 \dots : y = - 0.31253 \dots x + 1.02506 \dots

y = \frac{3}{50} x + \frac{7}{25} or y = - 0.31253 \dots x + 1.02506 \dots

\frac{d}{d x} (x^{2} (3 x - 4)^{3})

\int_{0}^{1} x^{2} e^{x} d x

y = \frac{16}{x ^{2} - 6 x - 55}, y = 0, x = - 3, x = 3

\frac{d}{d x} (\frac{x ^{3}}{( 1 - x ) ^{2}})

What is the tangent of f(x)= 1/(1+x^2),(2, 2/5) ?
The tangent of f(x)= 1/(1+x^2),(2, 2/5) is y= 3/50 x+7/25 \lor y=-0.31253…x+1.02506…