How do you find the critical points of y=x(1-x)^4? I know I need to take the derivative and I guess that's where I'm stuck. I think it's a product and a chain rule?

Feb 13, 2018

$\setminus$

$\text{The critical points are:} \setminus q \quad x = \frac{1}{5} , \setminus 1.$

Explanation:

$\setminus$

$\text{Yes, you are exactly right: it can be correctly thought of as}$
$\text{a product rule and chain rule problem.}$

$\text{Let's compute" \ y' ":}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad y \setminus = \setminus x {\left(1 - x\right)}^{4.}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad y \setminus = \setminus \left[x\right] \setminus \cdot \left[{\left(1 - x\right)}^{4}\right] .$

$\text{Product Rule:} \setminus q \quad \setminus q \quad \setminus \quad \setminus y ' \setminus = \setminus x \left[{\left(1 - x\right)}^{4}\right] ' + \left[x\right] ' \left[{\left(1 - x\right)}^{4}\right] .$

$\text{Chain Rule for first differentiated quantity:}$

$\setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus y ' \setminus = \setminus x \left[4 {\left(1 - x\right)}^{3} \left(1 - x\right) '\right] + 1 \setminus \cdot \left[{\left(1 - x\right)}^{4}\right] .$

$\text{Continuing (notice the -1 that appears -- important):}$

$\setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus y ' \setminus = \setminus x \left[4 {\left(1 - x\right)}^{3} \left(- 1\right)\right] + {\left(1 - x\right)}^{4.}$

$\text{Simplifying -- factor out lowest powers of same quantities:}$

$\setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus y ' \setminus = \setminus {\left(1 - x\right)}^{3} \left[x \left(4\right) \left(- 1\right) + {\left(1 - x\right)}^{1}\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus {\left(1 - x\right)}^{3} \left[x \left(- 4\right) + \left(1 - x\right)\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus {\left(1 - x\right)}^{3} \left[- 4 x + 1 - x\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus {\left(1 - x\right)}^{3} \left(1 - 5 x\right) .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus = \setminus \left(1 - 5 x\right) {\left(1 - x\right)}^{3.}$

$\text{Thus:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus y ' \setminus = \setminus \left(1 - 5 x\right) {\left(1 - x\right)}^{3.}$

$\setminus$

$\text{Now we can calculate the critical points.}$

$\text{Recall that the critical points are the points where the derivative}$
$\text{vanishes or is undefined.}$

$\text{So we solve:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad y ' = 0 \setminus \quad \setminus \quad \text{and" \quad \quad y' = "undefined}$

$\text{So we solve:}$

$\setminus \quad \left(1 - 5 x\right) {\left(1 - x\right)}^{3} = 0 \setminus q \quad \text{and" \qquad ( 1 - 5 x ) (1 - x )^3 = "undefined}$

$\text{In the second part of the previous, we note that}$
$\left(1 - 5 x\right) {\left(1 - x\right)}^{3} \setminus \setminus \text{is defined everywhere, so the second part}$
$\text{yields no solutions. So we continue solving only the first part:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \left(1 - 5 x\right) {\left(1 - x\right)}^{3} = 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 1 - 5 x = 0 \setminus q \quad \setminus q \quad {\left(1 - x\right)}^{3} = 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus 5 x = 1 \setminus q \quad \setminus q \quad 1 - x = 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad x = \frac{1}{5} \setminus q \quad \setminus q \quad x = 1$

$\text{These are our critical points.}$

$\setminus$

$\text{Summarizing:}$

$\text{The critical points of" \ \ y = x (1 - x )^4 \ \ "are:} \setminus q \quad x = \frac{1}{5} , \setminus 1.$