# Finding K such that this function has critical points

• Mar 10th 2011, 02:15 PM
youngb11
Finding K such that this function has critical points
Determine the conditions on parameter $\displaystyle k$ such that the function $\displaystyle \displaystyle f(x)=\frac{2x+4}{x^2-k^2}$ will have critical points.

Critical points are defined as where $\displaystyle f'(x)=0$ or when $\displaystyle f'(x)$ does not exist. I was thinking it would be $\displaystyle k>0$ but the back of the book has the answer as $\displaystyle k$ being between $\displaystyle -2$ and $\displaystyle 2$.

Was my reasoning correct or am I missing something? I'm not sure why $\displaystyle k$ has to be between $\displaystyle -2$ and $\displaystyle 2$.
• Mar 10th 2011, 02:25 PM
skeeter
Quote:

Originally Posted by youngb11
Determine the conditions on parameter $\displaystyle k$ such that the function $\displaystyle \displaystyle f(x)=\frac{2x+4}{x^2-k^2}$ will have critical points.

Critical points are defined as where $\displaystyle f'(x)=0$ or when $\displaystyle f'(x)$ does not exist. I was thinking it would be $\displaystyle k>0$ but the back of the book has the answer as $\displaystyle k$ being between $\displaystyle -2$ and $\displaystyle 2$.

Was my reasoning correct or am I missing something? I'm not sure why $\displaystyle k$ has to be between $\displaystyle -2$ and $\displaystyle 2$.

what did you get for a derivative?
• Mar 10th 2011, 02:46 PM
youngb11
Quote:

Originally Posted by skeeter
what did you get for a derivative?

Is this correct: $\displaystyle \displaystyle f'(x)=\frac{-2(x^2+4x+k^2)}{(x^2-k^2)^2}$ ?

Just looking at the denominator, wouldn't a positive k value produce points where the slope doesn't exist? Unless they aren't including asymptotes. In that case, how would I figure out where the slope equaled zero?
• Mar 10th 2011, 02:59 PM
skeeter
Quote:

Originally Posted by youngb11
Is this correct: $\displaystyle \displaystyle f'(x)=\frac{-2(x^2+4x+k^2)}{(x^2-k^2)^2}$ ?

Just looking at the denominator, wouldn't a positive k value produce points where the slope doesn't exist? Unless they aren't including asymptotes. In that case, how would I figure out where the slope equaled zero?

the derivative is correct.

note that critical values occur at x-values where the function is defined. $\displaystyle f(x)$ is undefined when $\displaystyle x = \pm k$ , so there are no values of $\displaystyle x$ where $\displaystyle f(x)$ is defined and $\displaystyle f'(x)$ is undefined.

that leaves the critical values where $\displaystyle f'(x) = 0$

$\displaystyle x^2 + 4x + k^2 = 0$

$\displaystyle x = \dfrac{-4 \pm \sqrt{4^2 - 4k^2}}{2}$
for critical values to exist, the discriminant, $\displaystyle (b^2-4ac) \ge 0$
$\displaystyle 4^2 - 4k^2 \ge 0$
solve this inequality for $\displaystyle k$