What are the points of inflection, if any, of #f(x)=x^7/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3 #?

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Answer 1 · 2015-12-09T14:40:41

The only inflection point is #(1,-43/105)#

#f(x)=x^7/42 - (3x^6)/10 + (6x^5)/5 - (4x^4)/3#

#f'(x) = x^6/6-(9x^5)/5+6x^4-(16x^3)/3#

#f''(x) = x^5-9x^4+24x^3-16x^2#

In order to determine where the concavity does change, we must first determine where it might change.

#f''(x)# is never undefined, so we need to solve #f''(x)=-#

#f''(x) = x^2(x^3-9x^2+24x-16)#

Use inspection or the rational zeros test to see that #1# is a zero of #x^3-9x^2+24x-16#.

Then use division or trial and error to factor

#x^3-9x^2+24x-16 = (x-1)(x^2-8x+16)#

# = (x-1)(x-4)^2#

This gives us:

#f''(x) = x^2(x-1)(x-4)^2#. So,

#f''(x) = 0# at #0, 1, "and " 4#.

But #0# and #4# are zeros of even multiplicity, so the sign of #f''(x)# does not change at those zeros.

The sign of #f''(x)# changes only at #x=1#.

( Note: this could also be determined by using a sign chart for #f''#.)

The inflection pint is the point #(1,f(1))#, so we do the arithmetic to find #f(1) = -43/105#, and

The inflection point is #(1,-43/105)#