We can clearly see from the illustration above that, for those intervals over which the graph of the function is concave down, the value of the first derivative is decreasing as x increases.
Let's consider the relationship between the graph of the function itself and that of its first derivative. The graph of ƒ( x) = 2 x 5 - 10 x 3 has both concave up and concave down regions We can illustrate the point by looking at the graphs of the polynomial function ƒ( x) = 2 x 5 - 10 x 3 and its first derivative once more. In general terms, a curved line segment is said to be concave up if it forms all or part of a bowl shape, and concave down if it forms all or part of a dome shape. If it does exist there, however, we may be able to more easily determine the nature of the critical point using the second derivative test.Īs we shall see, the second derivative at a point on a curve can tell us something about the concavity of the curve at that point. It must also be differentiable immediately to the left and right of the critical point to which the test is applied, although it is not necessary for the first derivative to exist at the critical point itself.
In order for the test to work, the function must be continuous over the defined interval. So, we know that we can determine the nature of a critical point by checking the sign of the first derivative on either side of it. We then just need to apply the first derivative test to each point to determine whether it is a local maximum, a local minimum, or neither. Since ƒ( x) is a polynomial function and is therefore "well behaved", we can find all of its critical points by solving ƒ′( x) = 0.
It also has an inflection point at x = 0, where the graph of ƒ′( x) touches the x axis at a single point, but does not intersect it. We can see that ƒ( x) has two local extrema, which occur where the graph of ƒ′( x) intersects the x axis. Here we see the graph of the polynomial function ƒ( x) = 2 x 5 - 10 x 3 and the graph of its first derivative function ƒ′( x) = 10 x 4 - 30 x 2. If ƒ′( x) has the same sign on both the left and the right of ( c, ƒ( c)), then ( c, ƒ( c)) is neither a local maximum nor a local minimum.Ĭonsider the illustration below. Suppose we have a function ƒ( x) that is continuous on the closed interval and differentiable on the open interval ( a, b), and that there exists a value c such that a 0 to the left of ( c, ƒ( c)) and ƒ′( x) 0 to the right of ( c, ƒ( c)), then ( c, ƒ( c)) is a local minimum. It's probably worth repeating our somewhat more formal statement describing the first derivative test. The slope is decreasing to the left of the local minimum, and increasing to the right of the local minimum. For a local minimum, the exact opposite occurs. We know that for a local maximum, the slope of a function (which is after all what the first derivative gives us) will be positive to the left of the local maximum (where the function is increasing in value), and negative to the right of the local maximum (where the function is decreasing in value). If you have read the page entitled "The First Derivative Test", you will know that we can use the first derivative to determine whether a specific critical point on the graph of a function is a local maximum, a local minimum, or neither.