What is the tangent of f(x)=(2x+1)^{1/2},\at x=2 ?

The tangent of f(x)=(2x+1)^{1/2},\at x=2 is y= 1/(5^{1/2)}x+5^{1/2}-2/(5^{1/2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=(2x+1)^{1/2},\at x=2

tangent of

f (x) = (2 x + 1)^{\frac{1}{2}}, at x = 2

y = \frac{1}{5 ^{\frac{1}{2}}} x + 5^{\frac{1}{2}} - \frac{2}{5 ^{\frac{1}{2}}}

Solution steps

Find the tangent point:

(2, 5^{\frac{1}{2}})

Find the slope of

f (x) = (2 x + 1)^{\frac{1}{2}} : \frac{df ( x )}{d x} = \frac{1}{( 2 x + 1 ) ^{\frac{1}{2}}}

Compute the slope of

f (x) = (2 x + 1)^{\frac{1}{2}}

x = 2 : m = \frac{1}{5 ^{\frac{1}{2}}}

Find the line with slope m=

\frac{1}{5 ^{\frac{1}{2}}}

and passing through

(2, 5^{\frac{1}{2}}) : y = \frac{1}{5 ^{\frac{1}{2}}} x + 5^{\frac{1}{2}} - \frac{2}{5 ^{\frac{1}{2}}}

y = \frac{1}{5 ^{\frac{1}{2}}} x + 5^{\frac{1}{2}} - \frac{2}{5 ^{\frac{1}{2}}}

\int \frac{x ^{4}}{x ^{4} - 1} d x

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

g (t) = t^{2} t

\frac{d}{d x} (2^{x} (π x^{2} + cos (x)))

x \to 9 - lim (\frac{x - 9}{∣ x - 9 ∣})

What is the tangent of f(x)=(2x+1)^{1/2},\at x=2 ?
The tangent of f(x)=(2x+1)^{1/2},\at x=2 is y= 1/(5^{1/2)}x+5^{1/2}-2/(5^{1/2)}