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Math Help - No idea of what to do here...

  1. #1
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    No idea of what to do here...

    Hello,

    I have no idea on how to solve the problem below. Do I take the intergral or do I set the given derivative equal to something?

    Thanks!



    Regards,
    Chris K.
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  2. #2
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    Hint: f'(x)=\frac{x^2-1}{1+\cos^2(x)}
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  3. #3
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    Quote Originally Posted by ckoeber View Post
    Hello,

    I have no idea on how to solve the problem below. Do I take the intergral or do I set the given derivative equal to something?

    Thanks!



    Regards,
    Chris K.
    See this formula:
    If f(x)=\int_a^x g(t)dt
    then f'(x)=g(x)
    By applying this formula, you will get the derivative.
    Now, It should be easy.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Hint: f'(x)=\frac{x^2-1}{1+\cos^2(x)}
    I figured that, but then I go ahead and set that to zero, right?

    I graphed the function and it is periodic so I am assuming that I need to find the answer in relations to pi.

    Am I correct?
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  5. #5
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    Quote Originally Posted by General View Post
    See this formula:
    If f(x)=\int_a^x g(t)dt
    then f'(x)=g(x)
    By applying this formula, you will get the derivative.
    Now, It should be easy.
    Thanks. Essentially I substitute the limit into the orginal function and work it out. But in this problem, as I mentioned to Plato, the graph is periodic.

    So the answer would be in relations to pi somehow, right?
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  6. #6
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    \begin{gathered}<br />
  f'( - 1) = 0,\;f'( - 1^ -  ) > 0,\;f'( - 1^ +  ) < 0 \Rightarrow \quad \max  \hfill \\<br />
  f'(1) = 0,\;f'(1^ -  ) < 0,\;f'(1^ +  ) > 0 \Rightarrow \quad \min  \hfill \\ <br />
\end{gathered}
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  7. #7
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    Quote Originally Posted by ckoeber View Post
    So the answer would be in relations to pi somehow, right?
    No, because we just need to find where it equals zero.

    f'(x)=\frac{x^2-1}{1+\cos^2(x)}=0

    \implies \frac{x^2-1}{1+\cos^2(x)} \cdot (1+\cos^2(x)) = 0 \cdot (1+\cos^2(x))

    \implies x^2-1 = 0
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  8. #8
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    Quote Originally Posted by drumist View Post
    No, because we just need to find where it equals zero.

    f'(x)=\frac{x^2-1}{1+\cos^2(x)}=0

    \implies \frac{x^2-1}{1+\cos^2(x)} \cdot (1+\cos^2(x)) = 0 \cdot (1+\cos^2(x))

    \implies x^2-1 = 0
    Thanks! This is what I was looking for! I thought I couldn't solve for x because the denominator was throwing me off.

    I feel dumb as all it took was simple arithmetic to solve

    Regards,
    Chris K.
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  9. #9
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    Quote Originally Posted by Plato View Post
    \begin{gathered}<br />
  f'( - 1) = 0,\;f'( - 1^ -  ) > 0,\;f'( - 1^ +  ) < 0 \Rightarrow \quad \max  \hfill \\<br />
  f'(1) = 0,\;f'(1^ -  ) < 0,\;f'(1^ +  ) > 0 \Rightarrow \quad \min  \hfill \\ <br />
\end{gathered}
    After another poster showed me how to solve I see how to came to this conclusion now.

    Thanks.
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