Solving functions with a quadratic denominator

[daz1990]

Hi,

I have a problem which I cannot get my head around. I have been trying all week for many hours at a time so I could really use some advice if possible.

The problem;

There are two values k1​and k2, such that k1<k2, ​ for which the function L(k) is not defined. What are the values of k1 and k2?

Given information;

f(k) = 12k / (6+5k) g(k) = 17k² - 17 L(k) = f(g(x))


The first step would be to find f(g(x)) which I understand is; 17( 12k / (6+5k) )² - 17 and therefore I believe this to be a quadratic denominator().

From here I do not know how to go about solving for the values of k1 and k2?

Any help would be gratefully received!! Please note this is a practice question I have been given.

Thank you in advance,

Darren

 

[Ishuda]

Hi,

I have a problem which I cannot get my head around. I have been trying all week for many hours at a time so I could really use some advice if possible.

The problem;

There are two values k1​and k2, such that k1<k2, ​ for which the function L(k) is not defined. What are the values of k1 and k2?

Given information;

f(k) = 12k / (6+5k) g(k) = 17k² - 17 L(k) = f(g(x))


The first step would be to find f(g(x)) which I understand is; 17( 12k / (6+5k) )² - 17 and therefore I believe this to be a quadratic denominator().

From here I do not know how to go about solving for the values of k1 and k2?

Any help would be gratefully received!! Please note this is a practice question I have been given.

Thank you in advance,

Darren

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

As a hint, f(g(k)) is found by replacing k by g(k) in the definition of f(k). In this example we would have
\(\displaystyle f(g(k))\, =\, \frac{12\, g(k)}{6\, +\, 5\, g(k)}\)
BTW: You are correct that L(k) will have a quadratic denominator so when that quadratic is zero, the function L(k) will be undefined.

 

[ksdhart]

Well, the first thing is that you've got the wrong expression. f(g(k)) means to evaluate the function f(k) where g(k) is the argument. What you've posted is g(f(k)).

\(\displaystyle f\left(g\left(k\right)\right)=\frac{12\cdot g\left(k\right)}{6+5\cdot g\left(k\right)}\)

\(\displaystyle g\left(f\left(k\right)\right)=17\cdot \left[f\left(k\right)\right]^2-17\)

Once you're working with the correct expression, you'll need to find places where the function doesn't exist. So, can you find any values where the denominator is 0?

 

[daz1990]

Sorry I meant L(k) = f(g(k)) not f(g(x))


I have just realised I have done g(f(x))!!

So f(g(x)) = 12(17k^2 - 17) / 6 + 5(17k^2 - 17)

I understand I need to find two values where the denominator is equal to 0. But I don't know how to do that?

Thanks.

 

[stapel]

Well, the first thing is that you've got the wrong expression. f(g(k)) means to evaluate the function f(k) where g(k) is the argument. What you've posted is g(f(k)).

I have just realised I have done g(f(x))!

No; as the helper pointed out, you had done g(f(k)). As well as the order of the functions, so also the correct variable is important.

f(g(x)) = 12(17k^2 - 17) / 6 + 5(17k^2 - 17)

I understand I need to find two values where the denominator is equal to 0. But I don't know how to do that?

It is odd that your class has arrived at rational function before simple quadratics were covered. Since we cannot here provide the missing lessons, please try online resources, such as here.

Once you have studied at least two lessons from the link, making sure that you've covered "factoring quadratics" (here) and "the Quadratic Formula" (here), please attempt the exercise. If you get stuck, please reply with a clear listing of all of your efforts, starting with setting the denominator equal to zero. Thank you!