What is the tangent of f(x)=3x^3-x^2+9,(2,29) ?

The tangent of f(x)=3x^3-x^2+9,(2,29) is y=32x-35

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+9,(2,29)

tangent of

f (x) = 3 x^{3} - x^{2} + 9, (2, 29)

y = 32 x - 35

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 3 x^{3} - x^{2} + 9

(2, 29) : m = 32

Find the line with slope m=

32

and passing through

(2, 29) : y = 32 x - 35

y = 32 x - 35

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

\frac{1}{( s ^{2} + 0.5 s + 1 )}

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{e ^{3 x}}{x ^{2}}

\frac{d y}{d x} = \frac{10 x ^{2}}{1 + x ^{3}}

x = 15 - y^{2}, x = y^{2} - 3

What is the tangent of f(x)=3x^3-x^2+9,(2,29) ?
The tangent of f(x)=3x^3-x^2+9,(2,29) is y=32x-35