Problem 22 - Calculus A New Horizon

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 \((-\infty, -\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, \infty)\); (b) Decreasing on \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\); (c) Concave up on \((0, \infty)\); (d) Concave down on \((-\infty, 0)\); (e) Inflection point at \(x = 0\).

Step 1: Find the First Derivative

To analyze where the function is increasing or decreasing, start by finding the first derivative of the function. Given \( f(x) = x e^{x^2} \), use the product rule, which states \( (uv)' = u'v + uv' \), where \( u = x \) and \( v = e^{x^2} \).Differentiating \( u = x \) gives \( u' = 1 \). Differentiating \( v = e^{x^2} \) requires the chain rule: \( v' = 2x e^{x^2} \).Applying the product rule:\[ f'(x) = 1 \cdot e^{x^2} + x \cdot 2x e^{x^2} = e^{x^2} + 2x^2 e^{x^2} = (1 + 2x^2) e^{x^2} \]

Step 2: Determine Intervals of Increase/Decrease

The function is increasing where its derivative is positive and decreasing where its derivative is negative. Solve \( f'(x) = (1 + 2x^2) e^{x^2} \geq 0 \).Since \( e^{x^2} > 0 \) for all \( x \), the sign of \( f'(x) \) depends on \( 1 + 2x^2 \). Set \( 1 + 2x^2 = 0 \), which gives \( x = \pm\frac{1}{\sqrt{2}} \).- The function is increasing on the intervals where \( 1 + 2x^2 > 0 \), i.e., \( (-\infty, -\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, \infty) \).- It is decreasing on the interval \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\).

Step 3: Find the Second Derivative

To determine concavity, find the second derivative. Start from \( f'(x) = (1 + 2x^2) e^{x^2} \).Using the product rule and chain rule again:\[ f''(x) = \left(4x e^{x^2} + 2x \cdot (1 + 2x^2) e^{x^2}\right) = \left(4x e^{x^2} + 2x + 4x^3\right) e^{x^2} = 2x(5 + 2x^2) e^{x^2} \]Simplifying gives:\[ f''(x) = 2x(5 + 2x^2) e^{x^2} \]

Step 4: Determine Intervals of Concavity

Analyze \( f''(x) = 2x(5 + 2x^2) e^{x^2} \) to find where \( f \) is concave up (\( f''(x) > 0 \)) and concave down (\( f''(x) < 0 \)).- \( f''(x) > 0 \) in intervals where \( x(5 + 2x^2) > 0 \). This occurs when: - \( x > 0 \) and \( 5 + 2x^2 > 0 \) (which is all \( x > 0 \)), giving the interval \( (0, \infty) \).- \( f''(x) < 0 \) occurs: - \( x < 0 \) which satisfies \( 5 + 2x^2 > 0 \), giving the interval \( (-\infty, 0) \).

Step 5: Identify Inflection Points

Inflection points occur where the concavity changes, which is when \( f''(x) = 0 \) or is undefined. Set \( 2x(5 + 2x^2) = 0 \).- Solving gives \( x = 0 \) as the potential inflection point, since \( 5 + 2x^2 = 5 + 2(0)^2 = 5 \), which is non-zero, the sign of \( f''(x) \) changes around \( x = 0 \). Thus, \( x = 0 \) is an inflection point.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, used to measure how a function changes as its input changes. They provide the rate of variation or slope of the function at any given point. For the function given in our exercise, \( f(x) = x e^{x^2} \), the derivative helps determine where the function is increasing or decreasing. - To find the derivative, we use the product rule, which is essential for differentiating products of two functions.- The product rule formula is \( (uv)' = u'v + uv' \), where \( u \) and \( v \) are functions of \( x \).- Here, \( u = x \) and \( v = e^{x^2} \). Differentiating \( u \) gives \( u' = 1 \), and differentiating \( v \) using the chain rule gives \( v' = 2x e^{x^2} \).- Applying the product rule results in the first derivative: \( f'(x) = (1 + 2x^2) e^{x^2} \).This derivative will be the key to understanding the behavior of the function.

Intervals of Increase and Decrease

Identifying the intervals on which a function is increasing or decreasing is crucial for understanding its overall behavior. To do this, we analyze the sign of the first derivative.- The function is increasing where its derivative \(f'(x)\) is positive and decreasing where \(f'(x)\) is negative.- Since the expression \( (1 + 2x^2) e^{x^2} \) contains \( e^{x^2} \), which is always positive, the sign of \(f'(x)\) depends on \( 1 + 2x^2 \).- Solving \(1 + 2x^2 = 0\) gives the critical points at \( x = \pm \frac{1}{\sqrt{2}} \).- The function increases on intervals \((-\infty, -\frac{1}{\sqrt{2}})\) and \((\frac{1}{\sqrt{2}}, \infty)\), and decreases on \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\).These intervals are essential for graphing the function and recognizing its slope characteristics.

Concavity and Inflection Points

Concavity tells us how the function curves, either upwards or downwards. It's closely linked with the second derivative of a function. Understanding concavity helps in identifying inflection points, which are points where the function's concavity changes.- The second derivative \(f''(x)\) provides information about concavity. For our function, \( f''(x) = 2x(5 + 2x^2) e^{x^2} \).- The function is concave up where \(f''(x) > 0\) and concave down where \(f''(x) < 0\).- Solving for \( f''(x) > 0 \), we find it's concave up in the interval \((0, \infty)\).- For \( f''(x) < 0 \), the interval is \((-\infty, 0)\).- Inflection points occur where the concavity changes, specifically at \( x = 0 \) for our exercise, marking a shift from concave down to concave up.Understanding these concepts helps in anticipating how the graph of the function will look and where significant changes occur.

Product Rule

The product rule is a differentiation rule used when dealing with functions that are products of two or more functions. It is vital in calculus for maintaining accuracy while differentiating complex product functions.- The formula is \( (uv)' = u'v + uv' \), which allows us to differentiate products like \( xe^{x^2} \) seamlessly.- In our exercise, the product rule played a key role in obtaining the first derivative.- It required differentiating \( u = x \) as \( u' = 1 \) and the chain rule application for \( v = e^{x^2} \), yielding \( v' = 2x e^{x^2} \).Through the product rule, we were able to grasp the changing dynamics of the function. This technique is central for analyzing and solving many calculus problems involving products of functions. Keep it handy as it opens up new insights into the behavior of functions like ours.