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

The tangent of f(x)=(e^2)/(x^2-2x),\at x=-2 is y=(3e^2)/(32)x+(5e^2)/(16)

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

tangent of

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

y = \frac{3 e ^{2}}{32} x + \frac{5 e ^{2}}{16}

Solution steps

Find the tangent point:

(- 2, \frac{e ^{2}}{8})

Find the slope of

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

Compute the slope of

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

x = - 2 : m = \frac{3 e ^{2}}{32}

Find the line with slope m=

\frac{3 e ^{2}}{32}

and passing through

(- 2, \frac{e ^{2}}{8}) : y = \frac{3 e ^{2}}{32} x + \frac{5 e ^{2}}{16}

y = \frac{3 e ^{2}}{32} x + \frac{5 e ^{2}}{16}

\int_{- 1}^{1} 2 e^{x} d x

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

x \to \infty lim (\frac{( e ^{x} )}{2 x ^{2}})

\int cos (3 x) sin^{2} (3 x) d x

f (t) = \frac{4 t}{4 + t}

What is the tangent of f(x)=(e^2)/(x^2-2x),\at x=-2 ?
The tangent of f(x)=(e^2)/(x^2-2x),\at x=-2 is y=(3e^2)/(32)x+(5e^2)/(16)