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How do you find the inflection points of #f(x)= 12x^5+45x^4-80x^3+6#?

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Jun 22, 2018

Answer:

Maximum: x=-4
Inflection: x=0
Minimum: x=1

Explanation:

In general, this is a hard problem for a quintic equation. But this particular quintic has some missing low order terms that help us a great deal.

To find the inflection and turning points of #f(x)#, we follow the usual procedure and set #f'(x)=0#.
#f(x)=12x^5+45x^4-80x^3+6# so
#f'(x)=60x^4+180x^3-240x^2#

Now this is a quartic equation. A full "solution by radicals" of the quartic is known, but it is long and complex (and painful to perform!). Fortunately, we can factorise this equation:

#f'(x)=60x^2(x^2+3x-4)=60x^2(x+4)(x-1)#.

So #f'(x)=0rArr x=#-4, 0, or 1.

To categorise these consider the second derivative:
#f''(x)=240x^3+540x^2-480x=60x(4x^2+9x-8)#

#f''(-4)=-4800<0#
#f''(0)=0#
#f''(1)=300>0#

So -4 is a maximum of the function and +1 is a minimum. To categorise 0 we need to take another derivative.

#f'''(x)=720x^2+1080x-480#, so #f'''(0)=-480!=0#

Thus 0 is a point of inflection.

We want to sanity check our answer by comparing to the graph of the function, but the maximum at -4 is off the scale compared to the other points of interest. Plot it twice, once zoomed in, once zoomed out:
graph{12x^5+45x^4-80x^3+6 [-5, 2, -20, 20]}
graph{12x^5+45x^4-80x^3+6 [-5, 2, -1000, 5000]}

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