# What is the equation of the tangent line of f(x) = e^(x)/(x^2-x at x=3?

May 27, 2018

$y = {e}^{3} / 36 x + {e}^{3} / 12$

#### Explanation:

$f \left(x\right) = {e}^{x} / \left({x}^{2} - x\right)$

D_f={AAx$\in$$\mathbb{R}$:x^2-x!=0}=(-oo,0)uu(0,1)uu(1,+oo)=RR-{0,1}

$f ' \left(x\right) = \left({e}^{x} / \left({x}^{2} - x\right)\right) ' = \frac{\left({e}^{x}\right) ' \left({x}^{2} - x\right) - {e}^{x} \left({x}^{2} - x\right) '}{{x}^{2} - x} ^ 2 =$

$\frac{{e}^{x} \left({x}^{2} - x\right) - {e}^{x} \left(2 x - 1\right)}{{x}^{2} - x} ^ 2 = \frac{{x}^{2} {e}^{x} - x {e}^{x} - 2 x {e}^{x} + {e}^{x}}{{x}^{2} - x} ^ 2 =$

$\frac{{x}^{2} {e}^{x} - 3 x {e}^{x} + {e}^{x}}{{x}^{2} - x} ^ 2$

For the equation of the tangent line at $A \left(3 , f \left(3\right)\right)$ we require the values

$f \left(3\right) = {e}^{3} / 6$

$f ' \left(3\right) = \frac{9 {e}^{3} - 9 {e}^{3} + {e}^{3}}{36} = {e}^{3} / 36$

The equation will be

$y - f \left(3\right) = f ' \left(3\right) \left(x - 3\right)$ $\iff$

$y - {e}^{3} / 6 = {e}^{3} / 36 \left(x - 3\right)$ $\iff$

$y - {e}^{3} / 6 = {e}^{3} / 36 x - \cancel{3} {e}^{3} / \cancel{36}$ $\iff$

$y = {e}^{3} / 36 x - {e}^{3} / 12 + {e}^{3} / 6$ $\iff$

$y = {e}^{3} / 36 x + {e}^{3} / 12$

and a graph