What is the tangent of f(x)=2xsqrt(x),\at x=1 ?

The tangent of f(x)=2xsqrt(x),\at x=1 is y=3x-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=2xsqrt(x),\at x=1

tangent of

f (x) = 2 x x, at x = 1

y = 3 x - 1

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

f (x) = 2 x x

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

y = 2 x, y = 2 x

y^{^{''}} - y^{^{'}} - 2 y = e^{x} + x

\frac{d}{d x} (\frac{1}{9} ln (ln (x)) + 3)

x^{3} y^{^{'}} = y

f (x) = (2 - 4 x^{2})^{5}, at x = 1

What is the tangent of f(x)=2xsqrt(x),\at x=1 ?
The tangent of f(x)=2xsqrt(x),\at x=1 is y=3x-1