What is the tangent of f(x)=sqrt(2x+3),(3,3) ?

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

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tangent of f(x)=sqrt(2x+3),(3,3)

tangent of

f (x) = 2 x + 3, (3, 3)

y = \frac{1}{3} x + 2

Solution steps

Find the slope of

f (x) = 2 x + 3 : \frac{df ( x )}{d x} = \frac{1}{2 x + 3}

Compute the slope of

f (x) = 2 x + 3

(3, 3) : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(3, 3) : y = \frac{1}{3} x + 2

y = \frac{1}{3} x + 2

x \to \infty lim (arctan (8 x))

\int_{1}^{\infty} \frac{9}{x ^{4}} d x

\frac{\partial}{\partial x} ((x^{4} - 1) (x^{2} + 1))

\frac{\partial}{\partial y} (y sin (x))

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