0
$\begingroup$

The function is:

$$y= 3x + \ln\left(\frac {3x - 4}{x - 1}\right)$$

After differentiating I got:

$$y' = 3 + \frac 1{3x^2 - 7x + 4}\;\;\;\;\; \;\;\; y''= - \frac {6x - 7}{(3x^2 - 7x + 4)^2}$$

If you can't find any critical points for $f'$ how do you determine where it is increasing/decreasing, and similarly for $y''$, where it is convex/concave?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Set $y'=0$ and note that you can multiply through by the denominator of that fraction. Hence you can find critical points. $\endgroup$ – anakhro Apr 27 '16 at 23:41
  • $\begingroup$ Well if $y'=0$ has no solutions then $y'$ can only change sign at points of discontinuity. To be clear, though, how do you define this for $1<x<\frac 43$? Do you leave it undefined? $\endgroup$ – lulu Apr 27 '16 at 23:41
  • $\begingroup$ The first derivative has no solutions, so just look at whether it is positive or negative to see if it the original function is increasing or decreasing (just plug in any viable test point into $y'$). The second derivative does have a solution, so you would try test points on either side to see if $y''$ is positive or negative, and on which intervals. $\endgroup$ – Dave Apr 27 '16 at 23:41
  • $\begingroup$ Be careful here...the derivative can change sign without ever being $0$, as it is undefined at least at $1$ and $\frac 43$. As I said in my earlier comment, the entire function is undefined between those values (unless you meant to take absolute value inside the log. $\endgroup$ – lulu Apr 27 '16 at 23:43
  • $\begingroup$ So the function is increasing in (-∞, 1) and (4/3, ∞)? $\endgroup$ – Phil Apr 27 '16 at 23:46
1
$\begingroup$

First, notice that x is not defined at $x=1$ and $x=\frac{4}{3}$.

Set $y'=0$

$3+\frac{1}{3x^2-7x+4}=0 \Rightarrow 3x^2-7x+4=-\frac{1}{3} \Rightarrow 3x^2-7x+\frac{13}{3}=0$

$(-7)^2-4*3*\frac{13}{3}=-3<0$, so it doesn't have a solution.

However, we were assuming $3x^2-7x+4 \neq 0$ before. Now consider the case that $3x^2-7x+4=0$, and we get $x=1$ or $x=\frac{4}{3}$. Neither of the two points fall in the range of x, so we don't have to consider these two cases.

We can see that $y'$ is always greater than $0$, which means y is monotonically increasing. And since it is monotonically increasing, it is neither convex nor concave.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.