What is the tangent of f(x)= 8/(sqrt(x)),\at x=36 ?

The tangent of f(x)= 8/(sqrt(x)),\at x=36 is y=-1/54 x+2

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tangent of f(x)= 8/(sqrt(x)),\at x=36

tangent of

f (x) = \frac{8}{x}, at x = 36

y = - \frac{1}{54} x + 2

Solution steps

Find the tangent point:

(36, \frac{4}{3})

Find the slope of

f (x) = \frac{8}{x} : \frac{df ( x )}{d x} = - \frac{4}{x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{8}{x}

x = 36 : m = - \frac{1}{54}

Find the line with slope m=

- \frac{1}{54}

and passing through

(36, \frac{4}{3}) : y = - \frac{1}{54} x + 2

y = - \frac{1}{54} x + 2

x \to 2 lim (3 x - 7)

(- e^{t}) + \frac{1}{2} (sin (t)) + \frac{1}{2} (cos (t))

x \to - \frac{π}{2} lim (sec (x))

\frac{d}{d x} ((sin (x) + x)^{2} + 7 cos (5 x))

- \frac{8 x}{x ^{2} + 1}

What is the tangent of f(x)= 8/(sqrt(x)),\at x=36 ?
The tangent of f(x)= 8/(sqrt(x)),\at x=36 is y=-1/54 x+2