What is the tangent of f(x)=sqrt(x+15),\at x=1 ?

The tangent of f(x)=sqrt(x+15),\at x=1 is y= 1/8 x+31/8

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tangent of f(x)=sqrt(x+15),\at x=1

tangent of

f (x) = x + 15, at x = 1

y = \frac{1}{8} x + \frac{31}{8}

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = x + 15 : \frac{df ( x )}{d x} = \frac{1}{2 x + 15}

Compute the slope of

f (x) = x + 15

x = 1 : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(1, 4) : y = \frac{1}{8} x + \frac{31}{8}

y = \frac{1}{8} x + \frac{31}{8}

(3 x^{2}) d x + (2 y - 4 x^{3} y^{- 1}) d y = 0

\int x sin^{2} (5 x) d x

\frac{d y}{d t} = 1 y

\int \frac{x + 2}{x + 1} d x

11 cos (- x)

What is the tangent of f(x)=sqrt(x+15),\at x=1 ?
The tangent of f(x)=sqrt(x+15),\at x=1 is y= 1/8 x+31/8