# What is the equation of the normal line of f(x)=((5-x)(4-x^2))/(x^3-1) at x=3?

Jan 6, 2018

$y = - 7.6 \overline{81} x + 22.66$

#### Explanation:

f(x)=((5−x)(4−x^2))/(x^3−1);quadquadquada=3

normal line: $\quad y = f \left(a\right) - \frac{1}{f ' \left(a\right)} \left(x - a\right)$

color(blue)(f(3)=)((5−3)(4−(3)^2))/(3^3−1)=(2*(-5))/(26)=color(blue)(-5/13)~~-0.384

Let: g(x)=(5−x)(4−x^2)
Then: $\textcolor{red}{g ' \left(x\right) =} - 1 \left(4 - {x}^{2}\right) + \left(5 - x\right) \cdot \left(- 2 x\right) = \textcolor{red}{- \left(4 - {x}^{2}\right) - 2 x \left(5 - x\right)}$

f'(x)=([-(4-x^2)-2x(5-x)]xx(x^3−1)-(5−x)(4−x^2)xx3x^2)/(x^3−1)^2

Don't bother to simplify. Just substitute x for 3

$f ' \left(3\right) = \frac{\left[- \left(4 - 9\right) - 6 \left(2\right)\right] \times \left(26\right) - \left(2\right) \left(- 5\right) \times 27}{26} ^ 2$

$f ' \left(3\right) = \frac{\left[+ 5 - 12\right] \times \left(26\right) + 10 \times 27}{26} ^ 2 = \frac{- 7 \times 26 + 270}{26} ^ 2$

$\textcolor{\mathmr{and} a n \ge}{f ' \left(3\right) =} \frac{- 182 + 270}{676} = \frac{88}{676} = \frac{44}{338} = \textcolor{\mathmr{and} a n \ge}{\frac{22}{169}} \approx 0.13$

$y = - \frac{5}{13} - \frac{1}{\frac{22}{169}} \left(x - 3\right)$

$y = - \frac{169}{22} x + \frac{6481}{286}$

$y = - 7.6 \overline{81} x + 22.66$