What is the tangent of f(x)=(20x)/(x^2-5),\at x=5 ?

The tangent of f(x)=(20x)/(x^2-5),\at x=5 is y=-3/2 x+25/2

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tangent of f(x)=(20x)/(x^2-5),\at x=5

tangent of

f (x) = \frac{20 x}{x ^{2} - 5}, at x = 5

y = - \frac{3}{2} x + \frac{25}{2}

Solution steps

Find the tangent point:

(5, 5)

Find the slope of

f (x) = \frac{20 x}{x ^{2} - 5} : \frac{df ( x )}{d x} = \frac{20 ( - x ^{2} - 5 )}{( x ^{2} - 5 ) ^{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{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(5, 5) : y = - \frac{3}{2} x + \frac{25}{2}

y = - \frac{3}{2} x + \frac{25}{2}

What is the tangent of f(x)=(20x)/(x^2-5),\at x=5 ?
The tangent of f(x)=(20x)/(x^2-5),\at x=5 is y=-3/2 x+25/2