What is the tangent of f(x)= 8/(sqrt(x)),\at x=25 ?

The tangent of f(x)= 8/(sqrt(x)),\at x=25 is y=-4/125 x+12/5

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tangent of f(x)= 8/(sqrt(x)),\at x=25

tangent of

f (x) = \frac{8}{x}, at x = 25

y = - \frac{4}{125} x + \frac{12}{5}

Solution steps

Find the tangent point:

(25, \frac{8}{5})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{4}{125}

Find the line with slope m=

- \frac{4}{125}

and passing through

(25, \frac{8}{5}) : y = - \frac{4}{125} x + \frac{12}{5}

y = - \frac{4}{125} x + \frac{12}{5}

x \to 0 lim (1 - 1 + x)

\int_{1}^{3} \frac{5}{x ^{2} + 4 x + 4} d x

\frac{d}{d x} (3 x^{2} e^{x} + e^{x} (x^{3} + 8))

\frac{d}{d y} (\frac{y ^{3}}{3})

x \to 1 - lim (∣ x ∣)

What is the tangent of f(x)= 8/(sqrt(x)),\at x=25 ?
The tangent of f(x)= 8/(sqrt(x)),\at x=25 is y=-4/125 x+12/5