How do you find the maximum, minimum, and inflection points for #h(x) = 7x^5 - 12x^3 + x#?

1 Answer
Jim H
Apr 3, 2015

Maximum and minimum:
Find critical numbers for #h#:

#h'(x)=35x^4-36x^2+1# which exists for all #x# and is #0# at,

#35x^4-36x^2+1 = 0#

#(35x^2-1)(x^2-1)=0#

Which has 4 solutions #x= +- 1/sqrt35#, #x= +-1#.

Now, test each critical number using either the first or second derivative tests for local extreme values.
Write your answer in your instructor's preferred form.

Inflection points:
#h''(x)=140x^3 - 72x = 4x(35x^2 - 18)#

So concavity could change at #x=0# and at #x= +- sqrt (18/35)#

A quick check of the sign (or an observation the these are all single zeros (i.e. not even multiplicity)) shows that the sign of #h''# and hance the concavity does indeed change at each of these values of #x#.

The inflection points are: #(0,0)#, #(sqrt (18/35), h(sqrt (18/35)))# and
#(-sqrt (18/35), h(-sqrt (18/35)))#

(If you are required to find the #y# values, save yourself some trouble by noticing that #h# is an odd function, so you only need to find #h(sqrt (18/35))# and then #h( - sqrt (18/35)) = - h(sqrt (18/35))# (which you just found).

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