What is the tangent of f(x)=sqrt(2x+1),\at x=4 ?

The tangent of f(x)=sqrt(2x+1),\at x=4 is y= 1/3 x+5/3

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tangent of f(x)=sqrt(2x+1),\at x=4

tangent of

f (x) = 2 x + 1, at x = 4

y = \frac{1}{3} x + \frac{5}{3}

Solution steps

Find the tangent point:

(4, 3)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(4, 3) : y = \frac{1}{3} x + \frac{5}{3}

y = \frac{1}{3} x + \frac{5}{3}

\frac{d}{d t} (\frac{y}{t} - (\frac{y}{t})^{2})

\int_{1}^{2} \frac{4}{x} - \frac{16}{x ^{2}} d x

\frac{\partial}{\partial y} (y - e^{- 2 x^{2} y})

\frac{\partial}{\partial x} (x (x - 2 y))

\int_{0}^{0.1} x arctan (2 x) d x

What is the tangent of f(x)=sqrt(2x+1),\at x=4 ?
The tangent of f(x)=sqrt(2x+1),\at x=4 is y= 1/3 x+5/3