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?

1 Answer
Feb 13, 2018

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"The critical points are:" \qquad x = 1/5, \ 1.

Explanation:

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"Yes, you are exactly right: it can be correctly thought of as"
"a product rule and chain rule problem."

"Let's compute" \ y' ":"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad y \ = \ x (1 - x )^4.

\qquad \qquad \qquad \qquad \quad :. \qquad \qquad \qquad \quad y \ = \ [ x ] \cdot [ (1 - x )^4 ].

"Product Rule:" \qquad \qquad \quad \ y' \ = \ x [ (1 - x )^4 ]' + [ x ]' [ (1 - x )^4 ].

"Chain Rule for first differentiated quantity:"

\qquad \qquad \quad \quad \ y' \ = \ x [ 4 (1 - x )^3 (1 - x )' ] + 1 \cdot [ (1 - x )^4 ].

"Continuing (notice the -1 that appears -- important):"

\qquad \qquad \quad \quad \ y' \ = \ x [ 4 (1 - x )^3 ( -1 ) ] + (1 - x )^4.

"Simplifying -- factor out lowest powers of same quantities:"

\qquad \qquad \quad \quad \ y' \ = \ (1 - x )^3 [x ( 4 ) ( -1 ) + (1 - x )^1 ]

\qquad \qquad \qquad \qquad \ \ = \ (1 - x )^3 [x ( -4 ) + (1 - x ) ]

\qquad \qquad \qquad \qquad \ \ = \ (1 - x )^3 [ -4 x + 1 - x ]

\qquad \qquad \qquad \qquad \ \ = \ (1 - x )^3 ( 1 - 5 x ).

\qquad \qquad \qquad \qquad \ \ = \ ( 1 - 5 x ) (1 - x )^3.

"Thus:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ y' \ = \ ( 1 - 5 x ) (1 - x )^3.

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"Now we can calculate the critical points."

"Recall that the critical points are the points where the derivative"
"vanishes or is undefined."

"So we solve:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad y' = 0 \quad \quad "and" \quad \quad y' = "undefined"

"So we solve:"

\quad ( 1 - 5 x ) (1 - x )^3 = 0 \qquad "and" \qquad ( 1 - 5 x ) (1 - x )^3 = "undefined"

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

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad ( 1 - 5 x ) (1 - x )^3 = 0

\qquad \qquad \qquad \qquad \qquad \qquad 1 - 5 x = 0 \qquad \qquad (1 - x )^3 = 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 5 x = 1 \qquad \qquad 1 - x = 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x = 1/5 \qquad \qquad x = 1

"These are our critical points."

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"Summarizing:"

"The critical points of" \ \ y = x (1 - x )^4 \ \ "are:" \qquad x = 1/5, \ 1.