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 April 14th, 2014, 09:54 AM #1 Newbie   Joined: Apr 2014 From: The Netherlands Posts: 7 Thanks: 2 Line intersections Edit: Oops! This is in the wrong section, maybe it better fits in pre-calc/calc. I'll just walk through the question as it is asked in the book I'm studying. Excuse me for grammar errors. Suppose we have the following function: $formdata=f(x)+=+\frac{6x}{x^2+++5}$ a) Calculate the extreme value of f(x) and give the range of f(x) So, we simply need to calculate f'(x) and set it to zero. Because I did answer a) correctly, I'll simply just get continue with posting f'(x) instead of deriving it step by step. $formdata=f'(x)+=+\frac{30-6x^2}{(x^2+5)^2}$ Now we need to set it to zero. $formdata=6x^2+=+30$ Which will yield: $formdata=x+=+\pm\sqrt{5}$ Putting these values in the original equation will result in the extreme values: $formdata=(-\sqrt{5},-\frac{3}{5}\sqrt{5})$ (minimum) and $formdata=(\sqrt{5},\frac{3}{5}\sqrt{5})$ (maximum). The range of f(x) will thus be $formdata=[-\frac{3}{5}\sqrt{5},\frac{3}{5}\sqrt{5}]$ b) Calculate (algebraically) for which a the equation f(x) = ax has exactly one unique solution. So, from this point on I have absolutely no idea what to do. Can anyone help me out? c) Calculate (algebraically) for which p the equation f(x) = 2/3x + p has three solutions. I thought that if I could get some help on b I could try and solve b and c in this topic :^). Last edited by jyrdo; April 14th, 2014 at 10:01 AM. Reason: Wrong section
 April 14th, 2014, 11:47 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra You need to find where $f(x) = ax$. That is the same as finding where $$g(x) = f(x) - ax = 0$$ So, try solving that for $x$, and see what value of $a$ gives a single (possibly repeated) solution in the real numbers.
April 14th, 2014, 12:13 PM   #3
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Quote:
 Originally Posted by v8archie You need to find where $f(x) = ax$. That is the same as finding where $$g(x) = f(x) - ax = 0$$ So, try solving that for $x$, and see what value of $a$ gives a single (possibly repeated) solution in the real numbers.
$formdata=\frac{6x}{x^2+5}+=+ax$

$formdata=x(ax^2+5a-6)+=+0$

I get that I have to find where $f(x) = ax$, but I don't think I'll find it the way I'm trying right now.

 April 15th, 2014, 03:04 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond $\displaystyle \frac{6x}{x^2+5}=ax$ $\displaystyle 6=ax^2+5a$ $\displaystyle ax^2=6-5a=0\Rightarrow a=\frac65$

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