 To solve an optimization problem, we need to use what's called the first derivative test. So first of all, remember that if a function is defined and differentiable over an interval, the endpoints always correspond to local extreme values. But what if you're not at an endpoint? If you're not at an endpoint, and the derivative is not equal to zero, then you're neither a local maximum nor a local minimum. And what that means is that if I want to find a local maximum or a local minimum, then I want the derivative to be zero or I want the derivative to not exist. And this leads to our definition of a critical point. The critical points of a function are the x-values for which the derivative is zero or the derivative does not exist. For example, suppose I have this function, and I want to find the critical points and determine whether they correspond to a local maximum, local minimum, neither or both. Since the critical points correspond to where the derivative is zero or fails to exist, my first task is to find the derivative. Now, if we were just finding the derivative, this would be a sufficient answer. However, because we do want to solve some equations with this, it'll probably be easiest to factor this at least partially. And in particular, we do want to factor out any common factors. So if we look at this expression, we see that both terms have a common factor of 3x minus 11 squared times 2x minus 7 to the third power. Removing this common factor gives us... and we can simplify this expression down to... and since this is a polynomial, the derivative is going to be defined for all x. So the only critical points will be where the derivative is equal to zero. So we'll find where the derivative is equal to zero, and because this is a product, I know that one of the factors has to be zero. And so that tells me the critical points will be at 11 thirds, 7 halves, and 151 42nd. Now, to decide whether a critical point corresponds to a local maximum, local minimum, neither or both, we need to find the sign of the derivative before and after each critical point. And we might make some observations. At each critical point, some of the factors are zero, and the sign of any individual factor can only change if the factor becomes zero or undefined. And what that means is that the critical points are the only places where the derivative could possibly change signs, and that's going to depend on the sign changes of the individual factors. To determine the sign of the derivative, let's determine the sign of each individual factor around each of the critical points. So we'll plot x equals 7 halves, 151 40 seconds, and 11 thirds, and we'll examine each of the factors, 2x minus 7 cubed, 42x minus 151, and 3x minus 11 squared. Now, note that our critical points have partitioned our interval into four parts and three boundaries. And what we'd like to do is we'd like to find the sign of the derivative in each of these seven pieces. So first, remember that these critical points, 7 halves, 151 40 seconds, and 11 thirds correspond to places where one of the factors is zero. So we know at x equals 7 halves, this factor 2x minus 7 to the third is zero, which will also mean our derivative is zero. And similarly, at x equals 151 over 40 second, the factor 42x minus 151 is zero, as is the derivative. And at x equals 11 thirds, the factor 3x minus 11 is zero, as is the derivative. Now let's take a look at each factor individually. If x is to the left of 7 halves, which is to say if x is less than 7 halves, then this factor 2x minus 7 to the third will be negative. And if x is greater than 7 halves, then 2x minus 7 to the third will be positive. Similarly, if x is less than 151 over 42, then this factor 42x minus 151 will be negative. And if x is greater than 151 over 42, this factor will be positive. And finally, if x is less than 11 thirds, then 3x minus 11 squared will be positive. And if x is greater than 11 thirds, 3x minus 11 squared will be positive again. And so now we know the signs of each factor that multiplies together to form f prime of x. So now we can find the sign of f prime of x itself. So if we're in this first region where x is less than 7 halves, then this first factor is negative. The second factor is negative. The third factor is positive. And f prime of x will be the product of two negatives and a positive. And this means that f prime of x will be positive. f prime of x will be zero at 7 halves. And then in this second interval, the first factor is positive. The second factor is negative. The third factor is positive. And so f prime of x will be negative. And then at x equals 151 over 42, f prime of x will be zero. In the third region, our sign analysis will tell us that f prime of x will be positive. And then in our last region, our sign analysis will tell us that f prime of x will be positive again. And since we know the sign of the derivative, we know a little bit about what the function does. And we can translate this into what the graph looks like. In order to do that, it's helpful to construct what we might call the stick figure version of the graph. So we're not going to worry about any sort of curvature or any other feature of the graph except for whether the graph is rising or falling. So in this first interval, our derivative is positive, so the graph is rising and our stick figure might look something like this. In the second interval, our derivative is negative, so f of x is decreasing and our graph is falling. So the graph looks something like this. In the third interval, the derivative is positive, so the graph looks like. And in the fourth interval, the derivative is positive, so the graph looks like. And so that tells us that at x equals 7 halves, we have a local maximum value. At x equals 151 over 42, we have a local minimum value. And x equals 11 thirds is the location of neither a maximum nor a minimum value.