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jazz1234567890

  • 2 years ago

Identify the relative maximum value for the function shown below. 7/x^2 + 5

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  1. richyw
    • 2 years ago
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    please typeset the function properly

  2. richyw
    • 2 years ago
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    Is this what you want (because this is what you typed) \[\frac{7}{x^2}+5\]

  3. jazz1234567890
    • 2 years ago
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    7 divided by x2+5

  4. jazz1234567890
    • 2 years ago
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    |dw:1352146335340:dw|

  5. richyw
    • 2 years ago
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    \[f(x)=\frac{7}{x^2+5}\]To maximize the function, take the derivative wrt x\[\frac{df}{dx}=\frac{d}{dx}\left(7(x^2+5)^{-1}\right)\]\[\frac{df}{dx}=7\frac{d}{dx}(x^2+5)^{-1}\]Use the chain rule...\[\frac{df}{dx}=7\left(-1(x^2+5)^{-2}\cdot \frac{d}{dx}(x^2+5)\right)\]\[\frac{df}{dx}=7\left(\frac{-2x}{\sqrt{x^2+5}}\right)\]\[\frac{df}{dx}=-\frac{14x}{\sqrt{x^2+5}}\]

  6. richyw
    • 2 years ago
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    where \(\frac{df}{dx}=0\) you have a critical point. So solve this for zero. It is easy to see that this occurs only where x=0. So that is your critical point. So find the value of \(f(x=0)\) Which is obviously \[f(0)=\frac{7}{5}\]Which is the maximum value of the function. Note that just because it is a critical point does not neccesarily mean that it is the maximum value, you must also check the behaviour at infinity (or the edges of a region if you are looking for the maximum value over a region). However in this problem it was given to you that there was a maximum, plus it is easy to see that \[\lim_{x\rightarrow \pm \infty} f(x)=0\] So you can be sure this is the maximum.

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