What is the tangent of f(x)=x^3+4x,(1,5) ?

The tangent of f(x)=x^3+4x,(1,5) is y=7x-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+4x,(1,5)

tangent of

f (x) = x^{3} + 4 x, (1, 5)

y = 7 x - 2

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{3} + 4 x

(1, 5) : m = 7

Find the line with slope m=

7

and passing through

(1, 5) : y = 7 x - 2

y = 7 x - 2

\frac{d y}{d x} = \frac{x y + y}{x + x y}

\int 6 sec^{6} (t) d t

\frac{\partial}{\partial t} (1 + t)

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

\int_{0}^{2} π y^{3} d y

What is the tangent of f(x)=x^3+4x,(1,5) ?
The tangent of f(x)=x^3+4x,(1,5) is y=7x-2