Determine the maxima and minima of $f(x)=-(x+7)^2+8$.

The first step is to graph the function. This will give you a visual idea of where the critical points, maxima, and minima are at. From the graph below it appears that there's only one maximum and it's between $x=-50$ and $x=0$.

Next, find the critical points. To do that you need the derivative. Using Rule 3 and the Chain Rule you know that $f\,'(x)=-2(x+7)$. So $f\,'(x)=0$ at $x=-7$ and the function has only one critical point.

The next step is to determine if there is a maximum, minimum, or neither at the critical point.

The quickest way is to use the second derivative test. Using our rules of differentiation we know that $f\,''(x)=-2$. The second derivative test says that there is a maximum if $f\,''(x)

We know there is a maximum at $x=-7$, but let's do the first derivative test anyway. Recall that from the first derivative test $f\,'(x)>0$ to the left of the critical point and $f\,'(x)

First, let's look to the left of the critical point. Since the critical point is at $x=-7$, this means that to the left of the critical point is $x\lt -7$. The way to do this is to apply High School algebra to this equation until you get $f\,'(x)$ on one side. Here's the algebra that does it:

\begin{eqnarray}x&\lt&-7\nonumber\\x+7&\lt&0\nonumber\\-2(x+7)&\gt&0\nonumber\\f\,'(x)&\gt&0\nonumber\end{eqnarray}

The important thing to remember is that when you multiply both sides by a negative number ($-2$ going from line 2 to line 3) the direction of $\lt$ is changed to $\gt$.

Now for $x$ to the right of $-7$ we have $x\gt-7$ and we can do the following algebra:

\begin{eqnarray}x&\gt&-7\nonumber\\x+7&\gt&0\nonumber\\-2(x+7)&\lt&0\nonumber\\f\,'(x)&\lt&0\nonumber\end{eqnarray}

So now that we've shown that $f\,'(x)\gt 0$ to the left of the critical point $x=-7$ and that $f\,'(x)\lt 0$ to the right, we've shown that there is a maximum at the critical point.

Both the first and second derivative tests confirm there's a maximum at $x=-7$.