What is the tangent of f(x)=3x^3-x^2+7,\at x=2 ?

The tangent of f(x)=3x^3-x^2+7,\at x=2 is y=32x-37

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=3x^3-x^2+7,\at x=2

tangent of

f (x) = 3 x^{3} - x^{2} + 7, at x = 2

y = 32 x - 37

Solution steps

Find the tangent point:

(2, 27)

Find the slope of

f (x) = 3 x^{3} - x^{2} + 7 : \frac{df ( x )}{d x} = 9 x^{2} - 2 x

Compute the slope of

f (x) = 3 x^{3} - x^{2} + 7

x = 2 : m = 32

Find the line with slope m=

32

and passing through

(2, 27) : y = 32 x - 37

y = 32 x - 37

6 x - x^{2}

\int_{1}^{5} \frac{x}{x - 1} d x

\int \frac{16 x ^{2}}{( x ^{2} + 9 ) ^{5}} d x

\int - 2 csc^{2} (x) cot (x) d x

\frac{d}{d x} (\frac{5 x}{x + 3})

What is the tangent of f(x)=3x^3-x^2+7,\at x=2 ?
The tangent of f(x)=3x^3-x^2+7,\at x=2 is y=32x-37