What is the tangent of f(x)=(4x)/(x^2-5),\at x=3 ?

The tangent of f(x)=(4x)/(x^2-5),\at x=3 is y=-7/2 x+27/2

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tangent of f(x)=(4x)/(x^2-5),\at x=3

tangent of

f (x) = \frac{4 x}{x ^{2} - 5}, at x = 3

y = - \frac{7}{2} x + \frac{27}{2}

Solution steps

Find the tangent point:

(3, 3)

Find the slope of

f (x) = \frac{4 x}{x ^{2} - 5} : \frac{df ( x )}{d x} = \frac{4 ( - x ^{2} - 5 )}{( x ^{2} - 5 ) ^{2}}

Compute the slope of

f (x) = \frac{4 x}{x ^{2} - 5}

x = 3 : m = - \frac{7}{2}

Find the line with slope m=

- \frac{7}{2}

and passing through

(3, 3) : y = - \frac{7}{2} x + \frac{27}{2}

y = - \frac{7}{2} x + \frac{27}{2}

\int \frac{1}{x x ^{2} - 9} d x

\int_{0}^{2} 1

\int (3 x^{5} - 4)^{2} \cdot 5 x^{4} d x

y^{^{''}} + 2 \cdot 2.5 \cdot 4 \cdot y^{^{'}} + 16 y = 0

\int \frac{1}{3 + 5 cos ( x )} d x

What is the tangent of f(x)=(4x)/(x^2-5),\at x=3 ?
The tangent of f(x)=(4x)/(x^2-5),\at x=3 is y=-7/2 x+27/2