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Thread: Increasing and Decreasing Functions

  1. #1
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    Increasing and Decreasing Functions

    Hi, I am stuck on this problem, the instructions indicate to find the intervals of increase and decrease for the given function.

    h(u) = square root of 9-u^2 The square root is over the entire term.

    We were told to differentiate first, and then to factor to find the zeros. When I try to differentiate I am getting 3-(1/2)X^-(3/2) This doesn't look right so I need some clarification as to whether I am differentiating correctly.

    BTW The final answer should be:

    h(u) is increasing for -3< u < 0
    h(u) is decreasing for 0< u < 3
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  2. #2
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    Quote Originally Posted by kdogg121 View Post
    Hi, I am stuck on this problem, the instructions indicate to find the intervals of increase and decrease for the given function.

    h(u) = square root of 9-u^2 The square root is over the entire term.

    We were told to differentiate first, and then to factor to find the zeros. When I try to differentiate I am getting 3-(1/2)X^-(3/2) This doesn't look right so I need some clarification as to whether I am differentiating correctly.

    BTW The final answer should be:

    h(u) is increasing for -3< u < 0
    h(u) is decreasing for 0< u < 3
    $\displaystyle h(u) = \sqrt{9 - u^2}$.

    From the chain rule: $\displaystyle h^{'}(u) = \frac{1}{2} (9 - u^2)^{-1/2} \times (-2u) = -\frac{u}{\sqrt{9 - u^2}}$

    where I've used the index laws $\displaystyle \sqrt{a} = a^{1/2}$ and $\displaystyle a^{-1/2} = \frac{1}{a^{1/2}} = \frac{1}{\sqrt{a}}$.

    Important note: h(u) is only defined for $\displaystyle -3 \leq u \leq 3$ .....
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  3. #3
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    So was I correct then? 3-(1/2)U^-(1/2) is the derivative?

    Will this factor?
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    Quote Originally Posted by kdogg121 View Post
    So was I correct then? 3-(1/2)U^-(1/2) is the derivative?
    Unfortunately not, no.

    Are you familiar with the chain rule? Here it is:

    $\displaystyle f[g(x)]=f'[g(x)]g'(x)$

    This theorem should be memorized, if you haven't already. So, applying it here, we can see the benefits. Our initial function:

    $\displaystyle h(u)=\sqrt{9-u^2}$

    Let $\displaystyle g(u)=9-u^2$

    Then:

    $\displaystyle h(u)=\sqrt{g(x)}$

    Can you finish it up?
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  5. #5
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    Quote Originally Posted by kdogg121 View Post
    So was I correct then? 3-(1/2)U^-(1/2) is the derivative?

    Will this factor?
    Sorry, but I can't see how either of the answers you've come up with are the same as the correct answer (see my first reply) of

    $\displaystyle h^{'}(u) = -\frac{u}{\sqrt{9 - u^2}}$.

    The solution to $\displaystyle h^{'}(u) = 0$ is clearly u = 0, obtained by setting the numerator equal to zero. The factoring business you mentioned is completely irrelevant .......

    Therefore:

    1. $\displaystyle h^{'}(u) < 0$ when u > 0 AND $\displaystyle -3 < u < 3$ (recall the important note from my earlier reply and also look at where $\displaystyle h^{'}(u) < 0$ is undefined), that is, $\displaystyle 0 < u < 3$.

    2. $\displaystyle h^{'}(u) > 0$ when yada yada
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  6. #6
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    I figured it out with the chain rule finally, thanks guys.
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