What is the equation of the normal line of f(x)=(x-3)^3(x+2) at x=-1?

Jun 16, 2018

$y + 64 = \frac{1}{16} \left(x + 1\right)$

Explanation:

First, find the point on the curve that the normal line will pass through:

$f \left(- 1\right) = {\left(- 4\right)}^{3} \left(1\right) = - 64$

So the normal line passes through $\left(- 1 , - 64\right)$.

Now recall that the normal line is perpendicular to the line tangent to the curve. So, if we know the slope of the tangent line at this point, we take its opposite reciprocal to find the slope of the perpendicular normal line.

To find the slope of the tangent line we take the derivative of the function. We could expand ${\left(x - 3\right)}^{3} \left(x + 2\right)$ into a polynomial but it might be easier to just use the product rule, since it's given already in this factored form.

$f ' \left(x\right) = \left(\frac{d}{\mathrm{dx}} {\left(x - 3\right)}^{3}\right) \left(x + 2\right) + {\left(x - 3\right)}^{3} \left(\frac{d}{\mathrm{dx}} \left(x + 2\right)\right)$

We use the chain rule on $\frac{d}{\mathrm{dx}} {\left(x - 3\right)}^{3}$. Note $\frac{d}{\mathrm{dx}} \left(x + 2\right) = 1$.

$f ' \left(x\right) = 3 {\left(x - 3\right)}^{2} \left(\frac{d}{\mathrm{dx}} \left(x - 3\right)\right) \left(x + 2\right) + {\left(x - 3\right)}^{3} \left(1\right)$

Moreover, $\frac{d}{\mathrm{dx}} \left(x - 3\right) = 1$, so:

$f ' \left(x\right) = 3 {\left(x - 3\right)}^{2} \left(x + 2\right) + {\left(x - 3\right)}^{3}$

$f ' \left(x\right) = {\left(x - 3\right)}^{2} \left[3 \left(x + 2\right) + \left(x - 3\right)\right]$

$f ' \left(x\right) = {\left(x - 3\right)}^{2} \left(4 x + 3\right)$

Then the slope of the tangent line at $x = - 1$ is:

$f ' \left(- 1\right) = {\left(- 4\right)}^{2} \left(- 1\right) = - 16$

Thus, the slope of the normal line is $\frac{1}{16}$.

The line passing through $\left(- 1 , - 64\right)$ with slope $\frac{1}{16}$ is:

$y + 64 = \frac{1}{16} \left(x + 1\right)$