Problem 10 - Essential Calculus Early Transcendentals

Question

(a) Find the intervals on which \(f\) is increasing or decreasing. (b) Find the local maximum and minimum values of \(f .\) (c) Find the intervals of concavity and the inflection points. \(f(x)=x^{4} e^{-x}\)

Step-by-Step Solution

Answer

Increasing on (-∞, 4), decreasing on (4, ∞). Local max at x=4; concavity changes at x=0, 2, 6.

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we need to find the first derivative of the function and analyze it. Given the function is \(f(x) = x^4 e^{-x}\), we apply the product rule. Let \(u = x^4\) and \(v = e^{-x}\). Then \(u' = 4x^3\) and \(v' = -e^{-x}\). The derivative is:\[f'(x) = u'v + uv' = 4x^3 e^{-x} - x^4 e^{-x} = e^{-x}(4x^3 - x^4)\].

Step 2: Find Critical Points

Set \(f'(x) = 0\) to find the critical points where the function may change from increasing to decreasing or vice versa. Solve:\[e^{-x}(4x^3 - x^4) = 0\].Since \(e^{-x}\) is never zero, solve \(4x^3 - x^4 = 0\):\[x^3(4 - x) = 0\].The solutions are \(x = 0\) and \(x = 4\).

Step 3: Determine Increasing/Decreasing Intervals

Analyze the sign of \(f'(x) = e^{-x}(4x^3 - x^4)\) in the intervals \((-\infty, 0)\), \((0, 4)\), and \((4, \infty)\):- For \(x < 0\), \(f'(x) > 0\) (since both \(e^{-x}\) and \(4x^3 - x^4\) are positive), function is increasing.- For \(0 < x < 4\), \(f'(x) > 0\) (as \(4x^3 \) dominates over \(x^4\)), function is increasing.- For \(x > 4\), \(f'(x) < 0\), function is decreasing.

Step 4: Identify Local Maxima and Minima

Using the critical points and nature of the function:- At \(x = 0\), \(f'(x)\) changes from positive to negative, indicating a local minimum.- At \(x = 4\), \(f'(x)\) changes from positive to negative, indicating a local maximum.

Step 5: Find the Second Derivative

To find concavity and inflection points, we need the second derivative. Differentiate \(f'(x) = e^{-x}(4x^3 - x^4)\) using the product rule again:\[f''(x) = e^{-x}[(4x^3 - x^4)' - (4x^3 - x^4)] = e^{-x}[(12x^2 - 4x^3) - 4x^3 + x^4]\].Simplifying, we have:\[f''(x) = e^{-x}(12x^2 - 8x^3 + x^4)\].

Step 6: Find Inflection Points and Concavity Intervals

Find where \(f''(x) = 0\):Solve \(e^{-x}(x^4 - 8x^3 + 12x^2) = 0\) which simplifies to:\[x^2(x^2 - 8x + 12) = 0\].Factor to get \(x^2(x-2)(x-6) = 0\), yielding \(x = 0, 2, 6\).Test intervals \((-\infty, 0)\), \((0, 2)\), \((2, 6)\), \((6, \infty)\) for concavity:- \(x < 0\) and \(x > 6\), \(f''(x) > 0\), function is concave up.- \(0 < x < 2\) and \(2 < x < 6\), \(f''(x) < 0\), function is concave down.Thus, inflection points occur at \(x = 0, 2, 6\).

Key Concepts

Increasing and Decreasing Intervals

To determine where the function is increasing or decreasing, we first need to calculate its first derivative. For the function \(f(x) = x^4 e^{-x} \), we use the product rule to find the derivative. This gives us:

• The first derivative: \( f'(x) = e^{-x}(4x^3 - x^4) \)

Next, we need to test the interval signs of the derivative to see where the function is positive (increasing) or negative (decreasing). Here's how we do it:

• On the interval \((-\infty, 0)\), \(f'(x) > 0\), meaning the function is increasing.
• On the interval \((0, 4)\), \(f'(x) > 0\), meaning the function continues to increase.
• On the interval \((4, \infty)\), \(f'(x) < 0\), indicating the function decreases.

By finding intervals where the derivative changes sign, we uncover where the function is smooth and climbing or falling over its domain.

Critical Points

Critical points occur where the first derivative is zero or undefined, possibly indicating a change in direction for the function. For \(f(x) = x^4 e^{-x} \), we set the first derivative equal to zero:

• \( f'(x) = e^{-x}(4x^3 - x^4) = 0 \).

Since \(e^{-x}\) is never zero, we solve \(4x^3 - x^4 = 0 \):

• This factors to \(x^3(4 - x) = 0\), giving solutions \(x = 0\) and \(x = 4\).

These critical points help determine where the function achieves local minimums or maximums. Assessing the sign change in the first derivative around these points reveals:

• At \(x = 0\), the derivative doesn't change sign, meaning no extremum.
• At \(x = 4\), \(f'(x)\) changes from positive to negative, indicating a local maximum.

Understanding critical points guides us in sketching a rough graph of a function.

Concavity and Inflection Points

Concavity describes the curvature of the function. The second derivative tells us about that curvature. For our function, we differentiate \(f'(x) = e^{-x}(4x^3 - x^4)\) to find

• The second derivative: \( f''(x) = e^{-x}(12x^2 - 8x^3 + x^4)\)

The sign of this second derivative

• Positive \(f''(x) > 0\) indicates concave up.
• Negative \(f''(x) < 0\) indicates concave down.

Inflection points occur where concavity changes, found by setting \(f''(x) = 0\):

• We solve \(x^2(x-2)(x-6) = 0\), leading to \(x = 0, 2, 6\).

By testing these intervals, we deduce:

• \(x < 0\) and \(x > 6\), where the function is concave up.
• \(0 < x < 2\) and \(2 < x < 6\), where it is concave down.

Inflection points occur at \(x = 0, 2, 6\), marking transitions in the graph's curvature.