Problem 29 - Calculus Early Transcendentals

Question

Use the given derivative to find all critical points of \(f\) and at each critical point determine whether a relative maximum, relative minimum, or neither occurs. Assume in each case that \(f\) is continuous everywhere. $$ f^{\prime}(x)=x e^{1-x^{2}} $$

Step-by-Step Solution

Answer

The critical point is at \(x = 0\), and it is a relative minimum.

Step 1: Identify Critical Points

Critical points occur where the derivative is zero or undefined. The given derivative is \( f'(x) = x e^{1-x^2} \). This derivative is defined for all \( x \), so we set it equal to zero: \[ x e^{1-x^2} = 0 \] This equation is satisfied if \( x = 0 \) because \( e^{1-x^2} eq 0 \) for any real \( x \). Thus, \( x = 0 \) is the only critical point.

Step 2: Determine the Second Derivative

To classify the critical point, we find the second derivative \( f''(x) \). Using the product rule for differentiation, \[ f''(x) = \frac{d}{dx}(x e^{1-x^2}) = e^{1-x^2} + x \cdot \frac{d}{dx}(e^{1-x^2}) \] We apply the chain rule to find \( \frac{d}{dx}(e^{1-x^2}) \): \[ \frac{d}{dx}(e^{1-x^2}) = e^{1-x^2} \cdot (-2x) = -2x e^{1-x^2} \] Thus, \[ f''(x) = e^{1-x^2} - 2x^2 e^{1-x^2} = (1 - 2x^2) e^{1-x^2} \].

Step 3: Evaluate the Second Derivative at the Critical Point

We substitute the critical point \( x = 0 \) into the second derivative: \[ f''(0) = (1 - 2(0)^2) e^{1-0^2} = e \] Since \( f''(0) > 0 \), the test indicates that \( x = 0 \) is a relative minimum.

Key Concepts

Derivative

In calculus, the derivative represents the rate at which a function is changing at any given point. It's like looking at the speedometer to see how fast you're going. More formally, for a function \( f(x) \), its derivative \( f'(x) \) describes the slope of the tangent line to the graph of the function. This is crucial for understanding where the function is increasing or decreasing.
Finding critical points involves taking the derivative of a given function and setting it to zero, as we did in the exercise with \( f'(x) = x e^{1-x^2} \). Critical points are those \( x \) values where the derivative is zero (which means the slope of the tangent is flat) or undefined.
In practice:

• Calculate the derivative of the function.
• Set the derivative equal to zero and solve for \( x \).
• Find where the derivative is undefined (if applicable).

Each of these steps helps in pinpointing where the potential peaks or valleys of the function lie.

Second Derivative Test

Once the critical points are found, the second derivative test is a way to classify them. It's like investigating further into a mystery to see whether you're looking at a hill or a valley. The second derivative, denoted as \( f''(x) \), tells us about the curvature of the function.
The process involves:

• Calculating the second derivative \( f''(x) \) of the function \( f(x) \).
• Substituting the critical points into \( f''(x) \).
• Interpreting the result:
  • If \( f''(x) > 0 \), the function is concave up at that point, indicating a relative minimum.
  • If \( f''(x) < 0 \), the function is concave down, indicating a relative maximum.
  • If \( f''(x) = 0 \), the test is inconclusive, and further investigation is needed.

In our case, a positive second derivative at \( x = 0 \) tells us we are at a relative minimum, affirming the point as a dip in the curve.

Relative Extrema

Relative extrema refer to the relative high and low points on a graph of a function. These are simply the local peaks (relative maxima) and valleys (relative minima) that occur in a given interval. Recognizing these can help in understanding the overall shape and behavior of the function.
Key ideas include:

• A relative maximum is where the function value is higher than all nearby points. Picture the top of a hill.
• A relative minimum is where the function value is lower than nearby points, similar to the bottom of a valley.
• These are determined by analyzing critical points using tests like the second derivative test.

In our exercise, determining that \( x = 0 \) is a relative minimum means that at this point, the function dips compared to values immediately around it.