Moselq8

2022-08-13

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:

$f(x)=h({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)dt).$

Find the intervals where f is decreasing and increasing, maxima and minima.

My try was this:

The derivative of f is given by the chain rule:

${f}^{\prime}(x)={h}^{\prime}({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)dt)\phi (\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})$

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: $(-\mathrm{\infty},-\surd 2)\cup (\surd 2,\mathrm{\infty})$ for the first case and $(-\surd 2,\surd 2)$ for the second one. Then (not sure of this part) ${h}^{\prime}({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)dt)>0$ and $\phi (\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})>0$ if $x\in (-\mathrm{\infty},-\surd 2)\cup (\surd 2,\mathrm{\infty}).$ Also if both h′ and $\phi $ are negative the product is positive, that's for $x\in (-\surd 2,\surd 2)$.

The case of the product being negative implies:

$x\in [(-\mathrm{\infty},-\surd 2)\cup (\surd 2,\mathrm{\infty})]\cap (-\surd 2,\surd 2)=[(-\mathrm{\infty},-\surd 2)\cap (-\surd 2,\surd 2)]\cup [(\surd 2,\mathrm{\infty})\cap (-\surd 2,\surd 2)]=\mathrm{\varnothing}.$

So the function is increasing in $(-\mathrm{\infty},-\surd 2),(-\surd 2,\surd 2),(\surd 2,\mathrm{\infty})$. So the function does not have maximum or minimum. Not sure of this but what do you think?

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:

$f(x)=h({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)dt).$

Find the intervals where f is decreasing and increasing, maxima and minima.

My try was this:

The derivative of f is given by the chain rule:

${f}^{\prime}(x)={h}^{\prime}({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)dt)\phi (\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})$

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: $(-\mathrm{\infty},-\surd 2)\cup (\surd 2,\mathrm{\infty})$ for the first case and $(-\surd 2,\surd 2)$ for the second one. Then (not sure of this part) ${h}^{\prime}({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)dt)>0$ and $\phi (\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})>0$ if $x\in (-\mathrm{\infty},-\surd 2)\cup (\surd 2,\mathrm{\infty}).$ Also if both h′ and $\phi $ are negative the product is positive, that's for $x\in (-\surd 2,\surd 2)$.

The case of the product being negative implies:

$x\in [(-\mathrm{\infty},-\surd 2)\cup (\surd 2,\mathrm{\infty})]\cap (-\surd 2,\surd 2)=[(-\mathrm{\infty},-\surd 2)\cap (-\surd 2,\surd 2)]\cup [(\surd 2,\mathrm{\infty})\cap (-\surd 2,\surd 2)]=\mathrm{\varnothing}.$

So the function is increasing in $(-\mathrm{\infty},-\surd 2),(-\surd 2,\surd 2),(\surd 2,\mathrm{\infty})$. So the function does not have maximum or minimum. Not sure of this but what do you think?

Payton Mcbride

Beginner2022-08-14Added 18 answers

Step 1

First of all, your derivative obtained via the Chain Rule doesn't look quite right. It should be ${f}^{\prime}(x)={h}^{\prime}({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)\phantom{\rule{thinmathspace}{0ex}}dt)\cdot \phi (\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})\cdot ({x}^{3}-x).$

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 $(\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})$, not multiplied by it.

Step 3

And third, you don't need to think much about ${h}^{\prime}(\cdots )$ and $\phi (\cdots )$ being positive or negative - simply because both are given to be always positive. Therefore, f′(x) has the same sign as $({x}^{3}-x)$, and so the only inequalities you need to solve are ${x}^{3}-x>0$ and ${x}^{3}-x<0$.

First of all, your derivative obtained via the Chain Rule doesn't look quite right. It should be ${f}^{\prime}(x)={h}^{\prime}({\int}_{0}^{\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2}}\phi (t)\phantom{\rule{thinmathspace}{0ex}}dt)\cdot \phi (\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})\cdot ({x}^{3}-x).$

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 $(\frac{{x}^{4}}{4}-\frac{{x}^{2}}{2})$, not multiplied by it.

Step 3

And third, you don't need to think much about ${h}^{\prime}(\cdots )$ and $\phi (\cdots )$ being positive or negative - simply because both are given to be always positive. Therefore, f′(x) has the same sign as $({x}^{3}-x)$, and so the only inequalities you need to solve are ${x}^{3}-x>0$ and ${x}^{3}-x<0$.

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