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

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

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

tangent of

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

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

Solution steps

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}

(3, 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}

x \to 7 lim (\frac{x + 2}{x - 7})

\int \frac{e ^{3 x} + 1}{e ^{2 x}} d x

\int_{0}^{1} \frac{x - 6}{x ^{2} - 5 x + 6} d x

\frac{d y}{d x} - e^{8 x + y} = 0

\int x \cdot sin (2 a x) d x

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