2
$\begingroup$

Does this function $$f(x)=\frac{2x+5}{3x+1}$$ not have any critical points? I have to find the intervals on which $f(x)$ is increasing or decreasing.

The first derivative of $f(x)$ is

$$f'(x)=-\frac{13}{(3x+1)^2}$$

which is undefined at $x=-1/3$ but that is not in the domain of $f(x)$, so $x=-1/3$ cannot be a critical point,right?

And for what value of x is $f'(x)=0$?

I also have to find the inflection points of $f(x)$ to fnd the intervals on which $f(x)$ is concave up or down so I found the 2nd derivative of $f(x)$,

$$f''(x)=\frac{78}{(3x+1)^3}$$

but then I got stuck again.

$\endgroup$

1 Answer 1

3
$\begingroup$

Your work is fine, we don't have critical points and $f'(x)\neq 0$ on the domain.

Moreover, since $f''(x)=\frac{78}{(1+3x)^3}$ we have

  • for $x> -\frac13 \implies f''(x)>0$ and the function is convex (concave up)

  • for $x< -\frac13 \implies f''(x)<0$ and the function is concave (down)


As an alternative approach, note that

$$y=\frac{2x+5}{3x+1} =\frac23 \frac{3x+\frac{15}2}{3x+1}=\frac23 \frac{3x+1+\frac{13}2}{3x+1}=\frac23+\frac{13}3\frac1{3x+1}$$

and by $X=3x+1$ and $Y=y-\frac23$ the function is in the form

$$Y=\frac A X$$

which is a rectangular hyperbola and all the results follows.

$\endgroup$
2
  • $\begingroup$ Is x=-1/3 the inflection point? $\endgroup$
    – Rodoshi
    Commented Jul 2, 2023 at 5:34
  • 1
    $\begingroup$ No it’s not, since this point is not in the domain, a necessary condition is $f’’(x)=0$. $\endgroup$
    – user
    Commented Jul 2, 2023 at 5:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .