Where does an inflection point occur for \(y=x^{3}-3 x\) ? a) 2 b) 1 c) 0 d) \(-1\)

Understanding the first derivative test is crucial for identifying the peaks and valleys of a graph, which are commonly referred to as local maxima and minima. This test uses the derivative of the function to determine where the graph of that function changes direction.

When the derivative of a function changes sign from positive to negative, we have found a local maximum. Conversely, if the derivative changes sign from negative to positive, a local minimum is at hand. To apply the first derivative test, we follow these steps:

• Find the critical points by setting the first derivative equal to zero.
• Examine the sign of the first derivative before and after each critical point.
• Based on the sign changes, determine whether each critical point is a local maximum, a local minimum, or neither.

In our example with the function ( y=x^{3}-3x), after finding its first derivative as (3x^2-3), setting it equal to zero reveals the critical points, which in this case are the roots of the equation ( x^2-1=0) or ( x= d 1) and ( x=-1). By testing values around these critical points, we can identify the nature of these points using the first derivative test.

The second derivative test is another analytical tool, but this one is based on the curvature of the function's graph. While the first derivative measures the rate of change, the second derivative assesses how the rate of change itself is changing – essentially, it looks at the 'bend' of the graph.

Here’s how we can conduct the second derivative test:

• Take the second derivative of the function.
• Find the critical points by equating the first derivative to zero.
• Plug these critical points into the second derivative.

If the second derivative at a critical point is positive, this indicates that the function is concave up at that point, and we have a local minimum. If it's negative, the function is concave down, and we've found a local maximum. If the second derivative is zero, the test is inconclusive.

In our example, stepping through the obligate differentiation rendered the second derivative ( y''=6x). At the critical points ( x= d 1) and ( x=-1), the second derivative test can't be applied since those are not points of inflection in this case. However, the inflection point was found where the second derivative equals zero, at ( x=0), which is not a local extremum but a point where concavity changes.

Critical points are pivotal in the study of calculus because they're the key to understanding the overall behavior of a function. They occur where the first derivative is either zero or undefined, marking potential maximums, minimums, or inflection points.

To locate critical points, one has to:

• Find the derivative of the function.
• Set that derivative equal to zero and solve for the variable.
• Also, check where the derivative is undefined.

It's essential to note that not all critical points will lead to a local maximum or minimum. That's why further tests like the first and second derivative tests are employed to classify these points accurately.

In our exercise, the critical points found when setting the first derivative ( y'=3x^2-3) to zero at ( x= d 1) and ( x=-1) also require analysis to define their nature. While they are key to revealing local extrema, in the context of looking for an inflection point, our interest was in where the second derivative equals zero, which only occurred at ( x=0), making it the inflection point. Therefore, the critical point at ( x=0) for this particular function is significant as it represents a change in concavity and hence determines the inflection point.