Finding critical points and inflection points for $f(x)=\frac{2x+5}{3x+1}$

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.

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.

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