Find absolute extrema of $f(x)=\frac{14}{x+2}$ on $[0,\infty)$

Question

Find absolute extrema of $f(x)=\frac{14}{x+2}$ on $[0,\infty)$

$Solution:$

The question is asking to find the absolute extrema of $f(x)$ on $[0,\infty)$, which means to find the points on the graph of $f(x)$ where the largest and smallest values of the function occur.

The first step is gather all the potential values of $x$ where an absolute extrema can happen. These points are exactly the values of $x$ where $f'(x)=0$ and the points on the boundary of our domain. Since our domain is $[0,\infty)$, the only boundary point on it is when $x=0$.

Lets solve where $f'(x)=0$ so we can gather the rest of the points where an absolute extrema could possibly occur.

$f(x) = 14(x+2)^{-1}$

and so by the chain rule:

$f'(x) = -14(x+2)^{-2} (\frac{d}{dx}(x+2))$

$f'(x) = -14(x+2)^{-2} (\frac{d}{dx}x+\frac{d}{dx}2)$

$f'(x) = -14(x+2)^{-2} (1+0)$

$f'(x) = -14(x+2)^{-2}$

Cool. We want to know when this function equals zero:

$f'(x) = \frac{-14}{(x+2)^{2}}=0$

If we multiply both sides of this equation by $(x+2)^{2}$, it becomes clear that there are no solutions:

$(x+2)^{2}\frac{-14}{(x+2)^{2}}=0(x+2)^{2}=0$

$\rightarrow -14=0$.... Umm... Nope!!

Thus the only point where an absolute extrema could occur is at the boundary point $x=0$... But how do we know if this value is an absolute maximum or absolute minimum? The answer is that it is an absolute maximum. One way to see this is that if we plug in number inside our domain of $[0,\infty)$ into $f(x)$, it's going to be smaller than if we plug zero in, because if $x$ isn't zero we will be dividing by a bigger number.

Another way to see that it is an absolute maximum is to see that the value of $f'(x)$ on $[0,\infty)$ is always negative, so it's always decreasing, so the value of $f'(x)$ gets smaller as $x$ gets bigger, thus we want to pick the smallest number in the domain, which is $x=0.$

Therefore the largest value that $f(x)=\frac{14}{x+2}$ achieves on $[0,\infty)$ occurs at $x=0$, so now we just need to find the corresponding $y$ value: $f(0)=7$.

So the absolute maximum is $(0,7)$ and the absolute minimum DNE.

Answer 1

That looks correct, but you don't need to compute $f'$ to do it. In fact, if $x,y\in[0,\infty)$, then$$x>y\implies x+2>y+2\implies\frac{14}{x+2}<\frac{14}{y+2}.$$So, $f$ is strictly decreasing on $[0,\infty)$ and therefore

• the maximum is attained at $0$ (and it is equal to $7$);
• the minimum does not exist.