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.

Moselq8 2022-08-13 Answered
If h has positive derivative and φ 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 φ is continuous and positive. If:
f ( x ) = h ( 0 x 4 4 x 2 2 φ ( t ) d t ) .
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 ( x ) = h ( 0 x 4 4 x 2 2 φ ( t ) d t ) φ ( x 4 4 x 2 2 )
We need to analyze where is positive and negative. So I solved the inequalities:
x 4 4 x 2 2 > 0 x 4 4 x 2 2 < 0
That gives: ( , 2 ) ( 2 , ) for the first case and ( 2 , 2 ) for the second one. Then (not sure of this part) h ( 0 x 4 4 x 2 2 φ ( t ) d t ) > 0 and φ ( x 4 4 x 2 2 ) > 0 if x ( , 2 ) ( 2 , ) . Also if both h′ and φ are negative the product is positive, that's for x ( 2 , 2 ).
The case of the product being negative implies:
x [ ( , 2 ) ( 2 , ) ] ( 2 , 2 ) = [ ( , 2 ) ( 2 , 2 ) ] [ ( 2 , ) ( 2 , 2 ) ] = .
So the function is increasing in ( , 2 ) , ( 2 , 2 ) , ( 2 , ). So the function does not have maximum or minimum. Not sure of this but what do you think?
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Answers (1)

Payton Mcbride
Answered 2022-08-14 Author has 18 answers
Step 1
First of all, your derivative obtained via the Chain Rule doesn't look quite right. It should be f ( x ) = h ( 0 x 4 4 x 2 2 φ ( t ) d t ) φ ( x 4 4 x 2 2 ) ( 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 x 4 4 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 φ of ( x 4 4 x 2 2 ) , not multiplied by it.
Step 3
And third, you don't need to think much about h ( ) and φ ( ) 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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