Where does the graph of #y=(5x^4)-(x^5)# have an inflection point?

Question

Emily Ammerman · Accepted Answer

An inflection point is a location on the graph where concavity changes. We'll examine concavity by examining the second derivative's sign: #y=5x^4 - x^5#  #y'=20x^3 - 5x^4#  #y''=60x^2 -20 x^3 = 20x^2(3-x)#  Obviously #20x^2# is always positive, so the sign of #y''# is the same as the sign of #3-x#.
Which is positive for #x<3# and negative for #x>3#.  At #x=3# the concavity changes.   A point on the graph is an inflection point, so we require: when #x=3#, we get 
#y = 5(3^4)-3^5 = 5(3^4)-3(3^4)=2(3^4) = 2(81)=162# The point #(3, 162)# in the only inflection point for the graph of #y=5x^4 - x^5#

Paisley Johnson · Answer

undefined To find the inflection points of the function $ y = 5x^4 - x^5 $, we need to find where the second derivative changes sign or where the second derivative is zero. First, let's find the first derivative $ y' $: $$ y' = \frac{dy}{dx} = 20x^3 - 5x^4 $$ Next, let's find the second derivative $ y'' $: $$ y'' = \frac{d^2y}{dx^2} = 60x^2 - 20x^3 $$ Now, set $ y'' $ to zero and solve for $ x $: $$ 60x^2 - 20x^3 = 0 $$ $$ 20x^2(3 - x) = 0 $$ From this equation, $ x = 0 $ or $ x = 3 $. Now, test the sign of $ y'' $ around these values to determine the inflection points: For $ x < 0 $: Pick $ x = -1 $ (for instance): $$ y'' = 60(-1)^2 - 20(-1)^3 = 60 - 20 = 40 $$ Since $ y'' > 0 $, there's no inflection point to the left of $ x = 0 $. For $ 0 < x < 3 $: Pick $ x = 1 $ (for instance): $$ y'' = 60(1)^2 - 20(1)^3 = 60 - 20 = 40 $$ Since $ y'' > 0 $, there's no inflection point between $ x = 0 $ and $ x = 3 $. For $ x > 3 $: Pick $ x = 4 $ (for instance): $$ y'' = 60(4)^2 - 20(4)^3 = 960 - 320 = 640 $$ Since $ y'' > 0 $, there's no inflection point to the right of $ x = 3 $. Therefore, the function $ y = 5x^4 - x^5 $ has an inflection point at $ x = 3 $.