# If h is differentiable and has positive derivative that pass through (0, 0), and psi is continuous and positive. If: f(x)=h(int_0^{x^4/4-x^2/2} psi(t)dt). Find the intervals where f is decreasing and increasing, maxima and minima.

If h has positive derivative and $\phi$ is continuous and positive. Where is increasing and decreasing f
The problem goes specifically like this:
If h is differentiable and has positive derivative that pass through (0,0), and $\phi$ is continuous and positive. If:
$f\left(x\right)=h\left({\int }_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi \left(t\right)dt\right).$
Find the intervals where f is decreasing and increasing, maxima and minima.
The derivative of f is given by the chain rule:
${f}^{\prime }\left(x\right)={h}^{\prime }\left({\int }_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi \left(t\right)dt\right)\phi \left(\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}\right)$
We need to analyze where is positive and negative. So I solved the inequalities:
$\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}>0\wedge \frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}<0$
That gives: $\left(-\mathrm{\infty },-\surd 2\right)\cup \left(\surd 2,\mathrm{\infty }\right)$ for the first case and $\left(-\surd 2,\surd 2\right)$ for the second one. Then (not sure of this part) ${h}^{\prime }\left({\int }_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi \left(t\right)dt\right)>0$ and $\phi \left(\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}\right)>0$ if $x\in \left(-\mathrm{\infty },-\surd 2\right)\cup \left(\surd 2,\mathrm{\infty }\right).$ Also if both h′ and $\phi$ are negative the product is positive, that's for $x\in \left(-\surd 2,\surd 2\right)$.
The case of the product being negative implies:
$x\in \left[\left(-\mathrm{\infty },-\surd 2\right)\cup \left(\surd 2,\mathrm{\infty }\right)\right]\cap \left(-\surd 2,\surd 2\right)=\left[\left(-\mathrm{\infty },-\surd 2\right)\cap \left(-\surd 2,\surd 2\right)\right]\cup \left[\left(\surd 2,\mathrm{\infty }\right)\cap \left(-\surd 2,\surd 2\right)\right]=\mathrm{\varnothing }.$
So the function is increasing in $\left(-\mathrm{\infty },-\surd 2\right),\left(-\surd 2,\surd 2\right),\left(\surd 2,\mathrm{\infty }\right)$. So the function does not have maximum or minimum. Not sure of this but what do you think?
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Payton Mcbride
Step 1
First of all, your derivative obtained via the Chain Rule doesn't look quite right. It should be ${f}^{\prime }\left(x\right)={h}^{\prime }\left({\int }_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi \left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt\right)\cdot \phi \left(\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}\right)\cdot \left({x}^{3}-x\right).$
You were missing the last part of the Chain Rule when applied to integrals with variable upper limit.
Step 2
Second, for some reason you were solving inequalities with $\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}$ being either greater or less than zero, as if it's a factor in the derivative - but it isn't! You have $\phi$ of $\left(\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}\right)$, not multiplied by it.
Step 3
And third, you don't need to think much about ${h}^{\prime }\left(\cdots \right)$ and $\phi \left(\cdots \right)$ being positive or negative - simply because both are given to be always positive. Therefore, f′(x) has the same sign as $\left({x}^{3}-x\right)$, and so the only inequalities you need to solve are ${x}^{3}-x>0$ and ${x}^{3}-x<0$.