Finding K such that this function has critical points

**youngb11** · March 10th 2011, 02:15 PM

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

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

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

**skeeter** · March 10th 2011, 02:25 PM

Originally Posted by youngb11

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

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

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

what did you get for a derivative?

**youngb11** · March 10th 2011, 02:46 PM

Originally Posted by skeeter

what did you get for a derivative?

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

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?

**skeeter** · March 10th 2011, 02:59 PM

Originally Posted by youngb11

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

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. $f(x)$ is undefined when $x = \pm k$ , so there are no values of $x$ where $f(x)$ is defined and $f'(x)$ is undefined.

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

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

using the quadratic formula ...

$x = \dfrac{-4 \pm \sqrt{4^2 - 4k^2}}{2}$

for critical values to exist, the discriminant, $(b^2-4ac) \ge 0$

$4^2 - 4k^2 \ge 0$

solve this inequality for $k$

**youngb11** · March 10th 2011, 03:04 PM

Thanks a lot for the help! I was somewhat on the right track, but was just a little confused. Thanks again!

Thread: Finding K such that this function has critical points

LinkBack

Thread Tools

Display

Finding K such that this function has critical points

Similar Math Help Forum Discussions

Need Help finding critical points and Inflection points. (graphing)

Finding Critical points of function

Finding All Critical Points of f(x,y)

Finding critical points

Finding Critical Points

Search Tags