Problem 43 - Thomas Calculus

Question

Graph the functions in Exercises \(37-46 .\) What symmetries, if any, do the graphs have? Specify the intervals over which the function is increasing and the intervals where it is decreasing. $$ y=x^{3} / 8 $$

Step-by-Step Solution

Answer

The graph of \(y = \frac{x^3}{8}\) is symmetric about the origin and is increasing on the interval \((-\infty, \infty)\).

Step 1: Determine Symmetry

The given function is \(y = \frac{x^3}{8}\). To determine the symmetry, test for even and odd functions. If \(f(-x) = f(x)\), the function is even. If \(f(-x) = -f(x)\), the function is odd. Substitute \(-x\) for \(x\): \(y = \frac{(-x)^3}{8} = \frac{-x^3}{8} = -\frac{x^3}{8}\). Since \(f(-x) = -f(x)\), the function is odd, indicating rotational symmetry about the origin.

Step 2: Find Critical Points

To find intervals of increase and decrease, find the derivative \(y'\). Calculate \(y' = \frac{d}{dx}\left(\frac{x^3}{8}\right) = \frac{3x^2}{8}\). Critical points occur where \(y' = 0\), but \(\frac{3x^2}{8} = 0\) only when \(x = 0\). Thus, no separate intervals beside this critical point can be determined.

Step 3: Determine Increasing or Decreasing Intervals

Using the derivative \(y' = \frac{3x^2}{8}\), note that \(y' > 0\) for all \(x eq 0\). This implies the function is increasing for all real numbers \(x\), as the square of a number is always non-negative, and a positive constant multiplier ensures the derivative remains positive.

Step 4: Graph the Function

Plot the function by choosing a few values of \(x\), both positive and negative, and compute \(y = \frac{x^3}{8}\). For example, when \(x = 2\), \(y = 1\); when \(x = -2\), \(y = -1\). The graph is a smooth curve passing through the origin with rotational symmetry around it, continuously rising in both directions due to its increasing nature.

Key Concepts

Function Symmetry

When examining a cubic function like \(y = \frac{x^3}{8}\), symmetry plays a vital role in understanding its graphical behavior.
Cubic functions have potential symmetries such as even or odd.

• If a function is even, you will find that \(f(-x) = f(x)\), leading to symmetry about the y-axis.
• If a function is odd, like in our example, the condition \(f(-x) = -f(x)\) holds true and shows rotational symmetry about the origin.

To illustrate this, substitute \(-x\) into the function: \[y = \frac{(-x)^3}{8} = -\frac{x^3}{8}\]This transformation is clear since it mirrors the original function over the origin, indicating an odd function and reinforcing rotational symmetry about the origin.
This essentially means that if you rotate the graph by 180 degrees around the origin, it will look the same.

Critical Points

Critical points in the graph of a function indicate places where there is the potential for the function to change its behavior, such as moving from increasing to decreasing or vice-versa.
For cubic functions like \(y = \frac{x^3}{8}\), identify these points by finding where the derivative equals zero.Start by calculating the derivative:\[y' = \frac{d}{dx}\left(\frac{x^3}{8}\right) = \frac{3x^2}{8}\]Set the derivative equal to zero to find critical points:\[\frac{3x^2}{8} = 0 \]This equation only holds true when \(x = 0\). At this point, the derivative indicates that there is no change from increasing to decreasing because the derivative itself is not negative or positive before and after the point, showing no shift in behavior, which leads us to the conclusion that \(x = 0\) is the only critical point but does not designate a change in direction.

Increasing and Decreasing Intervals

Analyzing where a function increases or decreases can help us understand how its graph behaves across different segments of the x-axis.
For the function \(y = \frac{x^3}{8}\), the behavior is guided by its derivative.We already know \[y' = \frac{3x^2}{8}\]This derivative simplifies to a non-negative value for all \(x\), but specifically, it is positive for all \(xeq 0\). This positive derivative suggests that:

• The function is always increasing as there's never a segment where \(y'\) is negative.
• Despite having a critical point at \(x = 0\), there is no actual interval where the function decreases.

Thus, this function exhibits pure increase across all real numbers, without any diminishing stretches - a hallmark property that firmly shapes the continuous, monotonic ascent of cubic functions like this one.