What are the inflection points of f(x)?

#f(x)=1.5x^5-5x^4-20x^3+120x^2-2x-6#

1 Answer
Noah G
Mar 29, 2018

#(-2, 510)# is the only point of inflection.

Explanation:

We must first find the second derivative.

#f'(x) = 7.5x^4 - 20x^3 - 60x^2 + 240x - 2#
#f''(x) = 30x^3 - 60x^2 - 120x + 240#

Inflection points occur when #f''(x) = 0#.

#0 = 30x^3 - 60x^2 - 120x + 240#

Thankfully we can readily solve this by factoring.

#0 = 30x^2(x - 2) -120(x - 2)#

#0 = (30x^2 - 120)(x- 2)#

#x = 2 or x^2 = 4#

#x = +-2#

But before declaring these, we must test the sign of the second derivative between these points. An inflection point only occurs if the sign changes.

Consider the following table:

enter image source here

As you can see, the second derivative does equal #0# at #x = 2#, but it does not goes from positive to negative. Instead it stays positive. Therefore, we must reject #x = 2#.

However, #x= -2# does have a sign change, therefore it's acceptable.

Hopefully this helps!

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