Question

A function \(f(x)\) is given. (a) Find the possible points of inflection of \(f\). (b) Create a number line to determine the intervals on which \(f\) is concave up or concave down. \(f(x)=e^{-x^{2}}\)

Short answer

Inflection points at \(x = \pm \frac{1}{\sqrt{2}}\). Concave up on \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), concave down elsewhere.

Step-by-step solution

Find the First Derivative

To begin, we calculate the first derivative of the function \(f(x) = e^{-x^2}\) using the chain rule. The derivative of \(e^u\) with respect to \(x\) where \(u = -x^2\) is \(e^u \cdot \frac{du}{dx}\). Thus, \(f'(x) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2}\).

Find the Second Derivative

Next, we need to find the second derivative, \(f''(x)\). To do this, differentiate \(f'(x) = -2xe^{-x^2}\). Apply the product rule: \((uv)' = u'v + uv'\) where \(u = -2x\) and \(v = e^{-x^2}\). This gives \(f''(x) = (-2)e^{-x^2} + (-2x)(-2xe^{-x^2}) = -2e^{-x^2} + 4x^2e^{-x^2}\).

Simplify and Solve for Inflection Points

Simplify the expression for the second derivative to determine where it is zero: \(f''(x) = e^{-x^2}(-2 + 4x^2)\). Set \(f''(x) = 0\), noting that \(e^{-x^2} eq 0\), hence \(-2 + 4x^2 = 0\). Solving gives \(4x^2 = 2\) which simplifies to \(x^2 = \frac{1}{2}\), leading to \(x = \pm \frac{1}{\sqrt{2}}\). These points are candidates for inflection points.

Create the Number Line and Test Concavity

Create a number line with the critical points \(x = -\frac{1}{\sqrt{2}}\) and \(x = \frac{1}{\sqrt{2}}\). Choose test points in the intervals formed: \((-\infty, -\frac{1}{\sqrt{2}})\), \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), and \((\frac{1}{\sqrt{2}}, \infty)\). Calculate \(f''(x)\) at each test point to determine the sign, indicating concavity. The sign of \(-2 + 4x^2\) at each point determines concavity: \(-2\), \(0\), and \(2\) respectively.

Determine Concavity Intervals

From the sign analysis: - If \(f''(x) > 0\), \(f\) is concave up. - If \(f''(x) < 0\), \(f\) is concave down.Hence, \(f\) is concave up on \((-rac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) and concave down on \((-\infty, -\frac{1}{\sqrt{2}})\) and \((\frac{1}{\sqrt{2}}, \infty)\).

Key concepts

Concavity

Concavity describes the way a function curves and is crucial to understanding its shape. A function can be concave up or concave down:

• If a function is concave up on an interval, it means the derivative is increasing in that interval. Think of it as the shape of a cup. It can "hold water".
• If a function is concave down, the derivative is decreasing. Imagine it as an upside-down cup. It cannot hold water.

Finding where a function changes concavity involves locating inflection points. These are where the second derivative, if it exists, equals zero or does not exist.

To apply this to the example, the second derivative test was used. The function's second derivative was set to zero to find possible inflection points, leading to the discovery of intervals where the function changes concavity. The intervals determined were where the sign of the second derivative changes, providing insights on the function being concave up or down.

Derivatives

Understanding derivatives is key to mastering a wide range of mathematical concepts, including finding inflection points. A derivative represents the rate at which a function is changing at any given point.
It is essentially the slope of the function at a particular point on its curve.

In our example, the first derivative of the function, found using the chain rule, helped identify how the function itself behaves. Finding the first derivative, like in the case of the function \(f(x) = e^{-x^2}\), is the stepping stone to finding its second derivative, which tells us about concavity and possible inflection points.
Derivatives are used to find critical points, analyze the curve, and predict the function behavior, all of which are crucial skills in calculus.

Chain Rule

The chain rule is a critical technique in differentiation. It is especially useful when dealing with composite functions—those made up of two or more functions.
The rule states that to differentiate a composite function, take the derivative of the outer function and multiply it by the derivative of the inner function.

Consider our function \(f(x) = e^{-x^2}\). We apply the chain rule because it involves an exponential function with a complex exponent.
Here's how it works:

• Take the derivative of the outer function, which is an exponential function, first. It results in \(e^{-x^2}\).
• Next, multiply it by the derivative of the inner function \(-x^2\), which is \(-2x\).

This process provides the first derivative, essential for further analysis. The chain rule thus serves as a powerful tool, enabling you to work with complicated functions and simplify them for easier differentiation.

Product Rule

The product rule is another foundational concept in calculus, used specifically when differentiating products of two functions.
According to the product rule, if you have two functions multiplied together, \( u(x) \) and \( v(x) \), then the derivative of their product \(uv\) is \( u'v + uv' \).

Applied in our example, after using the chain rule to get the first derivative, we take it further to determine the second derivative of \(f(x) = -2xe^{-x^2}\) using the product rule:

• Consider \(u = -2x\) and \(v = e^{-x^2}\).
• Differentiate each separately to get \(u' = -2\) and apply the chain rule again for \(v'\).
• Plug these into the product rule formula to find the complete derivative.

Through the product rule, you can efficiently tackle expressions that would otherwise be complex, thereby revealing much about the function's concavity, as seen in the calculation of the second derivative.