What is the equation of the line normal to f(x)=x ^3-3x^2  at x=4?

Aug 1, 2017

$y = \frac{1}{24} x - \frac{23}{6}$

Explanation:

We have: $f \left(x\right) = {x}^{3} - 3 {x}^{2}$

First, let's find the $y$-coordinate corresponding to the given value of $x$:

$R i g h t a r r o w f \left(4\right) = {\left(4\right)}^{3} - 3 \cdot {\left(4\right)}^{2}$

$R i g h t a r r o w f \left(4\right) = 64 - 48$

$\therefore f \left(4\right) = 16$

So we must find the equation of the normal line at the point $\left(4 , 16\right)$.

Then, let's differentiate $f \left(x\right)$:

$R i g h t a r r o w f ' \left(x\right) = 3 {x}^{2} - 6 x$

For $x = 4$, the slope of the line is:

$R i g h t a r r o w f ' \left(4\right) = 3 \cdot {\left(4\right)}^{2} - 6 \cdot \left(4\right)$

$R i g h t a r r o w f ' \left(4\right) = 48 - 24$

$\therefore f ' \left(4\right) = 24$

The gradient of the normal line is the reciprocal of this value:

$R i g h t a r r o w \text{Gradient of normal line} = {\left(f ' \left(4\right)\right)}^{- 1}$

$\therefore \text{Gradient of normal line} = \frac{1}{24}$

Now, let's express the equation of the normal line in point-slope form:

$R i g h t a r r o w y - {y}_{1} = m \left(x - {x}_{1}\right)$

$R i g h t a r r o w y - 16 = \frac{1}{24} \left(x - 4\right)$

$R i g h t a r r o w y - 4 = \frac{1}{24} x - \frac{1}{6}$

$\therefore y = \frac{1}{24} x - \frac{23}{6}$

Therefore, the equation of the normal line is $y = \frac{1}{24} x - \frac{23}{6}$.