Problem 10 - Applied Calculus

Question

Describe in words the interesting features of the graph, including the location of the critical points and where the function is monotonic (that is, increasing or decreasing). Then use the derivative and algebra to explain the shape of the graph. $$ f(x)=x^{3}-6 x+1 $$

Step-by-Step Solution

Answer

The function has critical points at \(x = -\sqrt{2}\) and \(x = \sqrt{2}\), and a point of inflection at \(x = 0\). It's increasing for \(x < -\sqrt{2}\) and \(x > \sqrt{2}\), decreasing between \(-\sqrt{2}\) and \(\sqrt{2}\), and changes concavity at \(x = 0\).

Step 1: Find the Derivative

Calculate the first derivative of the function to find where the function is increasing or decreasing, as well as the critical points. The given function is \(f(x) = x^3 - 6x + 1\). The derivative is obtained by using basic rules of differentiation: \(f'(x) = 3x^2 - 6\).

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Set \(f'(x) = 0\): \[3x^2 - 6 = 0\] \[3(x^2 - 2) = 0\] \[x^2 = 2\] \[x = \pm\sqrt{2}\]. These are the critical points of the function.

Step 3: Determine Increasing or Decreasing Intervals

Use the sign of \(f'(x)\) to determine where the function is increasing or decreasing.\- For \(x < -\sqrt{2}\), pick \(x = -2\): \(f'(-2) = 3(-2)^2 - 6 = 6 > 0\), so \(f(x)\) is increasing.\- For \(-\sqrt{2} < x < \sqrt{2}\), pick \(x = 0\): \(f'(0) = -6 < 0\), so \(f(x)\) is decreasing.\- For \(x > \sqrt{2}\), pick \(x = 2\): \(f'(2) = 3(2)^2 - 6 = 6 > 0\), so \(f(x)\) is increasing.

Step 4: Analyze Concavity and Points of Inflection

Find the second derivative \(f''(x)\) to determine concavity and points of inflection: \(f''(x) = 6x\).\- For \(x < 0\), \(f''(x) < 0\), so the graph is concave down.\- For \(x > 0\), \(f''(x) > 0\), so the graph is concave up.\- The second derivative changes sign at \(x = 0\), indicating a point of inflection at \((0, f(0))\).

Step 5: Describe Graph Features

The graph of \(f(x) = x^3 - 6x + 1\) has critical points at \(x = \pm\sqrt{2}\), where it changes from increasing to decreasing and vice versa. The function is increasing for \(x < -\sqrt{2}\) and \(x > \sqrt{2}\), and decreasing between \(-\sqrt{2} < x < \sqrt{2}\). There's a point of inflection at \(x = 0\), with the graph changing from concave down to concave up.

Key Concepts

Derivative

Understanding derivatives is key to analyzing the behavior of functions. A derivative, often denoted as \( f'(x) \), measures how a function's value changes as the input changes. In simpler words, it helps determine the rate of change or the slope of the function at any given point. When you calculate the derivative of a function, you're essentially finding a new function that describes how the original function is changing.

For the function \( f(x) = x^3 - 6x + 1 \), the derivative is \( f'(x) = 3x^2 - 6 \). This derivative tells us how \( f(x) \) fluctuates as \( x \) varies. By setting \( f'(x) = 0 \), we find the critical points where the function might change its direction (from increasing to decreasing or vice versa).

The derivative is a powerful tool because it provides insights into the nature of the function, highlighting when it has peaks, troughs, or flat segments. Knowing where \( f'(x) \) is zero or undefined is crucial in understanding the graph's overall shape and behavior.

Monotonic Function

A function's monotonicity describes whether it is consistently increasing or decreasing over a certain interval.
This information is essential for understanding trends and patterns in data represented by the function.

To determine if a function is increasing or decreasing, we can look at the sign of its first derivative \( f'(x) \).
For example, if \( f'(x) > 0 \) in an interval, the function is increasing in that interval, meaning as \( x \) increases, \( f(x) \) also increases.
Conversely, if \( f'(x) < 0 \), the function is decreasing.

For the function \( f(x) = x^3 - 6x + 1 \), the derivative \( f'(x) = 3x^2 - 6 \) helps us identify intervals of monotonicity. We found that:

• For \( x < -\sqrt{2} \), the function is increasing.
• For \( -\sqrt{2} < x < \sqrt{2} \), the function is decreasing.
• For \( x > \sqrt{2} \), the function is again increasing.

This alternating behavior between increasing and decreasing indicates changes in the trend of the function's graph.

Concavity

Concavity gives us information about the curvature of a function, helping us understand whether the function resembles a bowl facing up or down at various intervals. This is important because it affects how the function's rate of change is increasing or decreasing over time.

The second derivative \( f''(x) \) is used to determine the concavity of a function:

• If \( f''(x) > 0 \), the function is concave up, like the shape of a smile.
• If \( f''(x) < 0 \), the function is concave down, like a frown.

For our function, \( f(x) = x^3 - 6x + 1 \), the second derivative is \( f''(x) = 6x \).
When \( x < 0 \), \( f''(x) < 0 \), indicating the function is concave down in this range. On the other hand, when \( x > 0 \), \( f''(x) > 0 \), so the function is concave up.

This knowledge helps ascertain how the function might bend, offering a complete picture of its graphical nature.

Point of Inflection

Points of inflection are spots on a graph where the concavity changes from upwards to downwards, or vice versa.
These points are important because they indicate where the curve starts to change its bending direction, giving clues about transitions in the function's behavior.

A point of inflection occurs where the second derivative changes its sign. For our function, \( f(x) = x^3 - 6x + 1 \), the second derivative is \( f''(x) = 6x \). The sign of \( f''(x) \) changes at \( x = 0 \). Therefore, \( x = 0 \) is a point of inflection.

At this point, not only does the concavity switch direction, but it also provides a sense that the graph is "smoothing out." These points are useful in understanding more complex behaviors and symmetries within graphs, adding depth to how we interpret them.