What is the tangent of f(x)=8+5x^2-2x^3,(1,11) ?

The tangent of f(x)=8+5x^2-2x^3,(1,11) is y=4x+7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=8+5x^2-2x^3,(1,11)

tangent of

f (x) = 8 + 5 x^{2} - 2 x^{3}, (1, 11)

y = 4 x + 7

Solution steps

Find the slope of

f (x) = 8 + 5 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 10 x

Compute the slope of

f (x) = 8 + 5 x^{2} - 2 x^{3}

(1, 11) : m = 4

Find the line with slope m=

4

and passing through

(1, 11) : y = 4 x + 7

y = 4 x + 7

f (x) = \frac{4}{x}, (4, 1)

\int_{0}^{l n (3)} 2 (3 - x e^{x}) d x

\frac{d}{d x} (x^{4} + x^{3} + x^{2} + x + 1)

\int x 3^{x^{2}} d x

f (x) = 2 - \frac{4}{x}

What is the tangent of f(x)=8+5x^2-2x^3,(1,11) ?
The tangent of f(x)=8+5x^2-2x^3,(1,11) is y=4x+7