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-   -   Line intersections (http://mymathforum.com/algebra/42886-line-intersections.html)

jyrdo April 14th, 2014 09:54 AM

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: http://www.forkosh.com/mathtex.cgi?f...x%5E2+%2B+5%7D

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.

http://www.forkosh.com/mathtex.cgi?f...%2B5%29%5E2%7D

Now we need to set it to zero.

http://www.forkosh.com/mathtex.cgi?f...=6x%5E2+%3D+30

Which will yield: http://www.forkosh.com/mathtex.cgi?f...%5Csqrt%7B5%7D

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

http://www.forkosh.com/mathtex.cgi?f...sqrt%7B5%7D%29 (minimum) and http://www.forkosh.com/mathtex.cgi?f...sqrt%7B5%7D%29 (maximum).

The range of f(x) will thus be http://www.forkosh.com/mathtex.cgi?f...sqrt%7B5%7D%5D

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 :^).

v8archie April 14th, 2014 11:47 AM

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.

jyrdo April 14th, 2014 12:13 PM

Quote:

Originally Posted by v8archie (Post 190717)
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.

http://www.forkosh.com/mathtex.cgi?f...%2B5%7D+%3D+ax

http://www.forkosh.com/mathtex.cgi?f...B5a-6%29+%3D+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.

greg1313 April 15th, 2014 03:04 AM

$\displaystyle \frac{6x}{x^2+5}=ax$

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

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


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