Problem 28 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=x e^{x^{2}}\)

Step-by-Step Solution

Answer

(a) Increasing on \((-\infty,\infty)\); (b) Never decreasing; (c) Concave up on \((0, \infty)\); (d) Concave down on \((-\infty, 0)\); (e) Inflection point at \(x = 0\).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first find the derivative \( f'(x) \). For \( f(x) = x e^{x^2} \), we use the product rule: \( f'(x) = \left(1\right)e^{x^2} + x \left( e^{x^2} \cdot 2x \right) = e^{x^2} + 2x^2 e^{x^2} = e^{x^2}(1 + 2x^2) \).

Step 2: Determine Intervals of Increase/Decrease

Set the derivative \( f'(x) \) equal to zero and solve: \( e^{x^2}(1 + 2x^2) = 0 \). The exponential term \( e^{x^2} \) is never zero, so \( 1 + 2x^2 = 0 \) gives no real solution. Thus, \( f'(x) \) is always positive, indicating that \( f(x) \) is increasing on \( (-\infty, \infty) \). Therefore, it is never decreasing.

Step 3: Find the Second Derivative for Concavity

Find the second derivative \( f''(x) \). Start with \( f'(x) = e^{x^2}(1 + 2x^2) \) and use the product rule: \( f''(x) = e^{x^2} \cdot d/dx(1 + 2x^2) + (1 + 2x^2) \cdot d/dx(e^{x^2}) = e^{x^2}(4x) + (1 + 2x^2)(2xe^{x^2}) = e^{x^2}(4x + 2x + 4x^3) = 2x e^{x^2} (3 + 2x^2) \).

Step 4: Determine Intervals of Concavity

To find concavity changes, solve \( f''(x) = 0 \): \( 2x e^{x^2} (3 + 2x^2) = 0 \). Since \( e^{x^2} \) isn't zero, we solve \( 2x(3 + 2x^2) = 0 \). This gives \( x = 0 \) or \( 3 + 2x^2 = 0 \). However, \( 3 + 2x^2 = 0 \) gives no real solution. So, the change occurs at \( x = 0 \). Test intervals around \( x = 0 \): for \( x < 0 \), \( f''(x) < 0 \) (concave down); for \( x > 0 \), \( f''(x) > 0 \) (concave up).

Step 5: Identify Inflection Points

Since concavity changes at \( x = 0 \), \( x = 0 \) is an inflection point. Therefore, the only inflection point is at \( x = 0 \).

Key Concepts

Derivative

In calculus, the derivative of a function gives us the rate at which the function is changing at any point. The derivative is a foundational concept that allows us to understand a function's behavior and motion. For a function like \( f(x) = x e^{x^2} \), we find the derivative using the product rule, which states that if a function \( f(x) = u(x) \, v(x) \), then the derivative is \( f'(x) = u'(x) \, v(x) + u(x) \, v'(x) \). Here:

• \( u(x) = x \) with derivative \( u'(x) = 1 \).
• \( v(x) = e^{x^2} \) where chain rule gives \( v'(x) = e^{x^2} \cdot 2x \).

Putting it all together, we derive that \( f'(x) = e^{x^2}(1 + 2x^2) \). This derivative tells us how the function's slope or steepness changes along the x-axis. Understanding derivatives helps us determine where functions increase or decrease, which is crucial for analyzing graphs.

Concavity

Concavity describes the direction a function curves. If a function curves upwards like a cup, it is concave up. If it curves downwards like a frown, it’s concave down. To determine concavity, we look at the second derivative \( f''(x) \), which tells us how the slope of the function changes:

• If \( f''(x) > 0 \), the function is concave up on that interval.
• If \( f''(x) < 0 \), the function is concave down.

For \( f(x) = x e^{x^2} \), the second derivative is \( f''(x) = 2x e^{x^2} (3 + 2x^2) \). By analyzing the sign of \( f''(x) \):- For \( x < 0 \), \( f''(x) < 0 \) indicating the function is concave down.- For \( x > 0 \), \( f''(x) > 0 \) indicating the function is concave up.Concavity helps us understand the "bowl" shape of the graph, crucial for predicting the behavior of functions beyond simple slopes.

Inflection Points

Inflection points are where a function changes its concavity. It is a point on the graph where the curve shifts from concave up to concave down, or vice versa. To find inflection points, we set the second derivative \( f''(x) \) to zero and solve for \( x \). For \( f(x) = x e^{x^2} \), we have:- \( 2x e^{x^2} (3 + 2x^2) = 0 \).Since \( e^{x^2} \) never equals zero, we solve:

• \( 2x = 0 \), giving us \( x = 0 \). This is the point of inflection.

Analyzing the change of sign in \( f''(x) \) around \( x = 0 \) confirms it's an inflection point, as the concavity goes from down to up. Recognizing inflection points is essential for understanding the nature of a curve and predicting its graphical transition points.

Increasing and Decreasing Intervals

Understanding whether a function is increasing or decreasing helps us describe its overall trend. A function is increasing on intervals where its derivative \( f'(x) \) is positive and decreasing where \( f'(x) \) is negative.For the function \( f(x) = x e^{x^2} \), we determined that:

• \( f'(x) = e^{x^2}(1 + 2x^2) \) is always positive, as exponential functions are positive and \( 1+2x^2 \) is never negative.

This means \( f(x) \) is increasing on the entire interval \( (-\infty, \infty) \) and never decreases. This indicates a constantly rising graph. Identifying these intervals aids in predicting the overall behavior and trend of a function's graph.