## jazz1234567890 Group Title Identify the relative maximum value for the function shown below. 7/x^2 + 5 one year ago one year ago

1. richyw Group Title

2. richyw Group Title

Is this what you want (because this is what you typed) $\frac{7}{x^2}+5$

3. jazz1234567890 Group Title

7 divided by x2+5

4. jazz1234567890 Group Title

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5. richyw Group Title

$f(x)=\frac{7}{x^2+5}$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 Group Title

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.