What is the tangent of sqrt(3x+7),\at x=3 ?

The tangent of sqrt(3x+7),\at x=3 is y= 3/8 x+23/8

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tangent of sqrt(3x+7),\at x=3

tangent of

3 x + 7, at x = 3

y = \frac{3}{8} x + \frac{23}{8}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

y = 3 x + 7 : \frac{d y}{d x} = \frac{3}{2 3 x + 7}

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{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(3, 4) : y = \frac{3}{8} x + \frac{23}{8}

y = \frac{3}{8} x + \frac{23}{8}

x \to 2 lim (6 x + 1)

\int_{- \infty}^{- 1} \frac{1}{x ^{2}} d x

\int \frac{x ^{2} + 1}{x} \cdot ln (x) d x

\frac{3}{2 x + 4} + 9 x^{2}

x \to 0 + lim ((1 - 2 x)^{\frac{3}{x}})

What is the tangent of sqrt(3x+7),\at x=3 ?
The tangent of sqrt(3x+7),\at x=3 is y= 3/8 x+23/8