Question

In Exercises, find all relative extrema of the function. $$ h(x)=2(x-3)^{3} $$

Short answer

The given function \( h(x) = 2(x - 3)^3 \) has no relative extrema.

Step-by-step solution

Find the first derivative

Take the derivative of the function with respect to \(x\). The first derivative of \( h(x) = 2(x - 3)^3 \) is \( h'(x) = 6(x - 3)^2 \).

Find the critical points

To find the critical points, set the first derivative equal to zero and solve for \(x\), that is: \n \( 6(x - 3)^2 = 0 \). This simplifies to \( x - 3 = 0 \), which shows that \( x = 3 \) is the only critical point.

Find the second derivative and evaluate at the critical points

Now find the second derivative \( h''(x) = 12(x - 3) \). Now, evaluate this at the critical point \( x = 3 \). Substituting \( x = 3 \) into the equation we get \( h''(3) = 0 \).

Determine if the critical point is a relative maximum, minimum, or neither

Since the second derivative test is inconclusive at \( x = 3 \), there are no relative extrema.

Key concepts

Critical Points

Understanding the concept of critical points is foundational to analyzing the behavior of functions. Imagine hiking on a smooth hill; critical points are those flat spots where the slope of your path is level, so you're neither climbing up nor going down. Mathematically, we find these points on a graph of a function where its slope is zero or where the slope is undefined because the function takes a sharp turn.

In relation to our function, \( h(x) = 2(x - 3)^3 \), we calculated the first derivative to recognize where these level spots could be: \( h'(x) = 6(x - 3)^2 \). By setting this derivative equal to zero, we determined that \( x = 3 \) is our only critical point. At \( x = 3 \), the derivative does not change, implying a potential extremum or a point of inflection. This brings us to the question: is it a peak, a valley, or just a flat stretch? That's where the first and second derivative tests come into play.

First Derivative Test

The first derivative test is like dipping our toes into the water to gauge temperature—it's our initial check to understand the behavior around critical points. With this test, we observe the sign of the derivative (positive or negative) before and after the critical point to see whether the function is increasing or decreasing. We interpret this sign change to determine if a critical point is a local maximum or minimum.

For our example, after finding the critical point at \( x = 3 \), we would ideally check values just smaller and larger than 3 to see how the function acts. However, the derivative \( h'(x) = 6(x - 3)^2 \) is always non-negative for all values of \( x \), hence there's no sign change. The test tells us that the graph of our function is either always increasing or decreasing but doesn’t switch from one to the other—which correlates to neither a maximum nor a minimum at the critical point. However, since we're not seeing any sign change at all in our case, we cannot conclusively say what happens at \( x = 3 \) just yet.

Second Derivative Test

Think of the second derivative test as checking the curvature of a road while driving—you want to know if you're at the top of a hill (maximum), at the bottom of a valley (minimum), or on a stretch that's deceivingly flat (inflection point). The second derivative gives us an insight into the concavity of the function. A positive value indicates a concave up curve (smiley face), suggesting a local minimum, while a negative value indicates a concave down curve (frowny face), suggesting a local maximum.

When we ran this test on our function by taking the second derivative, \( h''(x) = 12(x - 3) \), and evaluating it at the critical point, we found that \( h''(3) = 0 \). This zero result is inconclusive; the test doesn't tell us whether \( x = 3 \) is a point of maximum, minimum, or inflection. That leaves us in a bit of a pickle as we're unable to determine the exact nature of critical point \( x = 3 \) solely from the second derivative. To fully classify this critical point, we might need to investigate further with other methods such as graphing or considering the context of the function within the problem.