What is the tangent of f(x)=sqrt(x-4),\at x=13 ?

The tangent of f(x)=sqrt(x-4),\at x=13 is y= 1/6 x+5/6

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tangent of f(x)=sqrt(x-4),\at x=13

tangent of

f (x) = x - 4, at x = 13

y = \frac{1}{6} x + \frac{5}{6}

Solution steps

Find the tangent point:

(13, 3)

Find the slope of

f (x) = x - 4 : \frac{df ( x )}{d x} = \frac{1}{2 x - 4}

Compute the slope of

f (x) = x - 4

x = 13 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(13, 3) : y = \frac{1}{6} x + \frac{5}{6}

y = \frac{1}{6} x + \frac{5}{6}

y^{^{'}} = - \frac{2 x}{y}

x \to 1 lim (∣ e^{x} - 1 ∣)

\int \frac{2}{3 x} d x

\frac{d y}{d x} = \frac{x}{y}, y (0) = - 7

y^{^{''}} + \frac{3}{x} y^{^{'}} = 0

What is the tangent of f(x)=sqrt(x-4),\at x=13 ?
The tangent of f(x)=sqrt(x-4),\at x=13 is y= 1/6 x+5/6