What is the tangent of f(x)=(x^3+5)/(x^2+1),\at x=-1 ?

The tangent of f(x)=(x^3+5)/(x^2+1),\at x=-1 is y= 7/2 x+11/2

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+5)/(x^2+1),\at x=-1

tangent of

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

y = \frac{7}{2} x + \frac{11}{2}

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

f (x) = \frac{x ^{3} + 5}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{x ^{4} + 3 x ^{2} - 10 x}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{x ^{3} + 5}{x ^{2} + 1}

x = - 1 : m = \frac{7}{2}

Find the line with slope m=

\frac{7}{2}

and passing through

(- 1, 2) : y = \frac{7}{2} x + \frac{11}{2}

y = \frac{7}{2} x + \frac{11}{2}

\int \frac{3 x - 3}{( x ^{2} - 2 x + 6 ) ^{2}} d x

\int 4 e^{2 x} d x

\frac{d}{d x} (\frac{1}{x ^{2} - 9})

\frac{d}{d x} ((x + e^{x}) (3 - x))

\int \frac{x ^{2}}{9 - x} d x

What is the tangent of f(x)=(x^3+5)/(x^2+1),\at x=-1 ?
The tangent of f(x)=(x^3+5)/(x^2+1),\at x=-1 is y= 7/2 x+11/2