1. ## Curve sketch analysis

Hi,

I am doing curve sketching in class and I don't understand an aspect of it. The curve to be sketched is $\displaystyle f(x)= \frac{x}{x-3}$

I understand the intercepts are (0,0) and (0,0), HA at y=1, VA at x=3 and domain all real except x=3 However, I am not sure how to find intervals of increase and decrease and relative extrema of the function

They have it listed as f'(x)= $\displaystyle \frac{-3}{(x-2)^2} < 0$ concluding that f decreases everywhere and no local extemas --> how do they arrive at these conclusions?

Also they have no POI for the graph after taking f''(x)=$\displaystyle \frac{6}{(x-3)^2}$ How do they make this conclusion?

2. ## Re: Curve sketch analysis

$\displaystyle f(x)=\frac{x}{x-3}$
To find the extrema of the function $\displaystyle f$ you have to calculate the first derivative which is
$\displaystyle f'(x)=\frac{-3}{(x-3)^2}$
and solve $\displaystyle f'(x)=0$. But $\displaystyle f'(x)$ can't never be equal to zero (why?) therefore there are no extremas and because the first derivative if always less then zero the function will decrease everywhere on his domain (if you don't know why take a look at the geometric meaning of the first derivative).

Your second derivative is incorrect (check it again). The conclusion is the same but in this case you're searching after (a) point(s) of inflection.