What is the tangent of f(x)=sqrt(3x+7),(3,4) ?

The tangent of f(x)=sqrt(3x+7),(3,4) is y= 3/8 x+23/8

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tangent of f(x)=sqrt(3x+7),(3,4)

tangent of

f (x) = 3 x + 7, (3, 4)

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

Solution steps

Find the slope of

f (x) = 3 x + 7 : \frac{df ( x )}{d x} = \frac{3}{2 3 x + 7}

Compute the slope of

f (x) = 3 x + 7

(3, 4) : 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}

\frac{1}{s ( s ^{2} + 2 s + 2 )}

(x - 2)^{2} (x + 4)

x^{\frac{9}{6}}, 2

n = 0 \sum \infty \frac{1}{sin ( n )}

x \to \frac{π}{6} lim (x sin (x))

What is the tangent of f(x)=sqrt(3x+7),(3,4) ?
The tangent of f(x)=sqrt(3x+7),(3,4) is y= 3/8 x+23/8