What is the tangent of f(x)=sqrt(x+8),(1,3) ?

The tangent of f(x)=sqrt(x+8),(1,3) is y= 1/6 x+17/6

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tangent of f(x)=sqrt(x+8),(1,3)

tangent of

f (x) = x + 8, (1, 3)

y = \frac{1}{6} x + \frac{17}{6}

Solution steps

Find the slope of

f (x) = x + 8 : \frac{df ( x )}{d x} = \frac{1}{2 x + 8}

Compute the slope of

f (x) = x + 8

(1, 3) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(1, 3) : y = \frac{1}{6} x + \frac{17}{6}

y = \frac{1}{6} x + \frac{17}{6}

n = 1 \sum \infty 3^{n} 5^{1 - n}

x \to \infty lim (\frac{x ^{\frac{1}{2}}}{x ^{\frac{2}{3}}})

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

x \to 2 lim (\frac{x}{x - 2})

(x + y)^{2} + (2 x y + x^{2} - 1) \frac{d y}{d x} = 0

What is the tangent of f(x)=sqrt(x+8),(1,3) ?
The tangent of f(x)=sqrt(x+8),(1,3) is y= 1/6 x+17/6