Find the second derivative implicitly with respect to x for x^4 + y^4 = 16
Find the second derivative implicitly with respect to x for x^4 + y^4 = 16

It is clear from the graph of $x^4$ that it should have a positive second derivative at 0 like the graph of $x^2$ but instead because it has 0 the test fails. So what is the fundamental problem with the test because of which it fails for this graph.

It is clear from the graph of $x^4$ that it should have a positive second derivative at 0 like the graph of $x^2$ but instead because it has 0 the test fails. So what is the fundamental problem with the test because of which it fails for this graph.

  • 2$\begingroup$ What “test” are you referring to? $\endgroup$ Jul 26, 2022 at 16:52
  • $\begingroup$ $x^4$ curves up slower than any $ax^2,a>0$ at $x=0$, so comparing its behavior to a quadratic function (which is basically what the second derivative test does) tells you nothing. $\endgroup$– IanJul 26, 2022 at 16:53
  • 1$\begingroup$ The second derivative test here has not given you any false information, so it has not “failed” (and does not have a fundamental problem). $\quad$ The higher-derivative test reveals that the point in question is a local minimum. $\endgroup$– ryangJul 26, 2022 at 17:18
  • $\begingroup$ If $f'(x)=0$ and $f”(x)\gt 0$ then we have a local minimum. The converse is not necessarily true. In fact, $f(x)=|x|$ has a local minimum at $x=0$ and it is not even differentiable at that point. $\endgroup$ Jul 26, 2022 at 17:26

1 Answer

You can try this way:

If the function is concave up at some point $x_0$, then it means the first derivative function $f'(x)$ is increasing at the neighborhood of $x_0$. So what you can do is to compare the value for $f'(x_0-\epsilon)$ and $f'(x_0+\epsilon)$.

For this problem, $f'(x)=4x^3$ and $x_0=0$, so you can compare $f'(-\epsilon)$ and $f'(+\epsilon)$.

$$-4\epsilon^3=f'(-\epsilon)<f'(+\epsilon)=4\epsilon^3$$

Since the first derivative is increasing at the neighborhood of $0$, so it is concave up there.

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