If h has positive derivative and φ is continuous and positive. Where is increasing and decreasing fThe 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 &gt; 0 ∧ x 4 4 − x 2 2 &lt; 0That 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 ) &gt; 0 and φ ( x 4 4 − x 2 2 ) &gt; 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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If h has positive derivative and   φ is continuous and positive. Where is increasing and decreasing fThe 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    &amp;gt;  0  ∧            x      4        4    −            x      2        2    &amp;lt;  0That 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    )    &amp;gt;  0 and   φ      (                  x        4            4        −                  x        2            2        )    &amp;gt;  0 if   x  ∈  (  −  ∞  ,  −      √    2  )  ∪  (      √    2  ,  ∞  )  . Also if both h′ and   φ are negative the product is positive, that&#039;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?

Payton Mcbride · Accepted Answer

Step 1First of all, your derivative obtained via the Chain Rule doesn&#039;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 2Second, for some reason you were solving inequalities with             x      4        4    −            x      2        2   being either greater or less than zero, as if it&#039;s a factor in the derivative - but it isn&#039;t! You have   φ of       (                  x        4            4        −                  x        2            2        )  , not multiplied by it.Step 3And third, you don&#039;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  &amp;gt;  0 and       x    3    −  x  &amp;lt;  0.

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.

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