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

The tangent of f(x)=x^3+4x-4,(1,1) is y=7x-6

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-4,(1,1)

tangent of

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

y = 7 x - 6

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 1) : m = 7

Find the line with slope m=

7

and passing through

(1, 1) : y = 7 x - 6

y = 7 x - 6

\int_{0}^{0.5} (1 - x^{2})^{0.5} d x

\int \frac{1}{a x} d x

\frac{d}{d x} (z + y^{3} + y + x)

x \to - 2 lim (x^{2} - 4 x + 10)

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

What is the tangent of f(x)=x^3+4x-4,(1,1) ?
The tangent of f(x)=x^3+4x-4,(1,1) is y=7x-6