How do you find the inflection points for the function f(x)=x^3+x?

1 Answer
Wataru
Oct 10, 2014

f(x)=x^3+x

By taking derivatives,

f'(x)=3x^2+1

f''(x)=6x=0 Rightarrow x=0,

which is the x-coordinate of a possible inflection point. (We still need to verify that f changes its concavity there.)

Use x=0 to split (-infty,\infty) into (-infty,0) and (0,infty).

Let us check the signs of f'' at sample points x=-1 and x=1 for the intervals, respectively.
(You may use any number on those intervals as sample points.)

f''(-1)=-6<0 Rightarrow f is concave downward on (-infty,0)

f''(1)=6>0 Rightarrow f is concave upward on (0,infty)

Since the above indicates that f changes its concavity at x=0, (0,f(0))=(0,0) is an inflection point of f.

I hope that this was helpful.

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