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

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

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

tangent of

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

y = \frac{1}{3} x + 3 - \frac{1}{3}

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

f (x) = 2 x + 1

x = 1 : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(1, 3) : y = \frac{1}{3} x + 3 - \frac{1}{3}

y = \frac{1}{3} x + 3 - \frac{1}{3}

\frac{d}{d x} (\frac{2 x + ∣ x ∣}{1 + ∣ x ∣})

\int \frac{9 x ^{2} - 8 x + 20}{x ^{3} + 4 x} d x

\frac{75 x}{x ^{2} + 100}

1 - 2 x

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

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