Consider $$ f(x)=x^4-3x^2+17 $$

To find the inflection points, I take the second derivative and determine its roots:

$$ f''(x)=(x^4-3x^2+17)''=(4x^3-6x)'=12x^2-6 $$

$$ 12x^2-6=0 \\ 12x^2=6 \\ x^2=\frac{1}{2} \\ x=\pm\sqrt{\frac{1}{2}}=\pm\frac{\sqrt{2}}{2} $$

This semms right to me. But the graph of f(x) looks like this: enter image description here Here, $$ x=\pm\frac{\sqrt{2}}{2} \approx\pm0.7$$ doesn't seem to have any consequence to me, shouldn't I get $$ x \approx 1.225 $$ for an inflection point? That looks like the point where f(x)'s graph changes from concave to convex...

  • 2
    $\begingroup$ Your x-values are right, but your graph is missing the x-axis, which makes it a little hard to estimate. It makes sense that there are two points of inflection, somewhere "between" the minimum and maximum of the graph. Ok I see the horizontal scale now, but I am not sure where you got that 1.225 from $\endgroup$ – imranfat Sep 21 '16 at 15:25
  • $\begingroup$ Does it? Then I am sorry, my question was thus pointless $\endgroup$ – bertalanp99 Sep 21 '16 at 15:28
  • $\begingroup$ Question is not pointless, your work is right because 0.707 is somewhere between 0 and 1.732 (location maximum). The 1.225 plays no role here $\endgroup$ – imranfat Sep 21 '16 at 15:30

The "$x \approx 1.225$" you're talking about sounds like $x=\sqrt{6}/2$, which is a minimum of $f(x)$ -- a place where the function changes from decreasing to increasing.

At $x=\sqrt{2}/2$, there is an inflection point -- a place where the function changes from concave down (and decreasing) to concave up (and decreasing).


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