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April 14th, 2014, 10:54 AM   #1
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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:

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.



Now we need to set it to zero.



Which will yield:

Putting these values in the original equation will result in the extreme values:

(minimum) and (maximum).

The range of f(x) will thus be

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 11:01 AM. Reason: Wrong section
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April 14th, 2014, 12:47 PM   #2
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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.
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April 14th, 2014, 01:13 PM   #3
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Quote:
Originally Posted by v8archie View Post
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.




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.
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April 15th, 2014, 04:04 AM   #4
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$\displaystyle \frac{6x}{x^2+5}=ax$

$\displaystyle 6=ax^2+5a$

$\displaystyle ax^2=6-5a=0\Rightarrow a=\frac65$
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