What are the points of inflection, if any, of #f(x)=4x^3 + 21x^2 - 294x +7 #?

1 Answer
Dec 7, 2017

#(7/2,-2373/4) and (-7,1722)# are points of minimum and maximum respectively, though none are at a point where inflection occurs, but the graph continues going in the same general direction.

Explanation:

An inflection occurs when #f'(x)=0#

To determine this, we must differentiate #f(x)dx#

#f(x)=4x^3 + 21x^2 - 294x +7#

#f'(x)=12x^2+42x-294#

#12x^2+42x-294=0#

#6x^2+21x-147=0#

Using the quadratic formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

We get:
#x=(-21+-sqrt(21^2-4(6*-147)))/(2(6))#

#x=(-21+-sqrt(441-4(-882)))/(12)#

#x=(-21+-sqrt(441+3528))/(12)#

#x=(-21+-sqrt(3969))/(12)#

#x=(-21+63)/(12)or(-21-63)/(12)#

#x=42/12or-84/12#

#x=7/2 or -7#

#f(7/2)=4(7/2)^3 + 21(7/2)^2 - 294(7/2)+7=-2373/4#

#f(-7)=4(-7)^3 + 21(-7)^2 - 294(-7)+7=1722#

The points of inflection are:
#(7/2,-2373/4) and (-7,1722)#

However, this only tells us if a point is either a maximum, minimum or inflection.

The second derivative can help us with this. If #f''(x)=0# it is inflection, #f''(x)>0# is a minimum, #f''(x)<0# is a maximum.

#f''(x)=24x+42#

#f''(7/2)=24(7/2)+42=126# (Therefore a minimum)

#f''(-7)=24(-7)+42=-126# (Therefore a maximum)

There no points of inflection as such, only minimum and maximum points.