Problem 14 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=2 x^{5}+15 x^{4}+30 x^{3}-42 $$

Step-by-Step Solution

Answer

Concave up on \((-\infty, -3)\) and \((0, \infty)\); concave down on \((-3, 0)\). Points of inflection: \(-3, 0\). Local max at \(x = -3\), local min at \(x = 0\).

Step 1: Find the First Derivative

To start with, we need to find the first derivative of the given function \( f(x) \). The function is \( f(x) = 2x^5 + 15x^4 + 30x^3 - 42 \). By applying the power rule, we get:\[ f'(x) = \frac{d}{dx}(2x^5 + 15x^4 + 30x^3 - 42) = 10x^4 + 60x^3 + 90x^2. \]

Step 2: Find the Second Derivative

Next, we find the second derivative, which helps in determining the concavity. The first derivative we obtained is \( 10x^4 + 60x^3 + 90x^2 \). Differentiating again, we find:\[ f''(x) = \frac{d}{dx}(10x^4 + 60x^3 + 90x^2) = 40x^3 + 180x^2 + 180x. \]

Step 3: Determine Concavity Intervals

Set the second derivative equal to zero to find critical points for concavity:\[ 40x^3 + 180x^2 + 180x = 0. \]Factor out the greatest common factor:\[ 40x(x^2 + 4.5x + 4.5) = 0. \]This gives \( x = 0 \) or solving \( x^2 + 4.5x + 4.5 = 0 \) using the quadratic formula gives\[ x = \frac{-4.5 \pm \sqrt{4.5^2 - 4 \times 4.5}}{2} \approx -3 \text{ and } -1.5. \]Test intervals around these values (\(-\infty, -3), (-3, -1.5), (-1.5, 0), (0, \infty\)) to determine the concavity on each interval.

Step 4: Identify Points of Inflection

Points where the concavity changes are points of inflection. These occur at \( x = -3, -1.5 \) and \( 0 \) since the sign of \( f''(x) \) changes at these points.

Step 5: Find Critical Points Using First Derivative

Find where the first derivative is zero or undefined to find critical points. Set \( f'(x) = 0 \):\[ 10x^4 + 60x^3 + 90x^2 = 0 \]Factor:\[ 10x^2(x^2 + 6x + 9) = 0 \] giving \( x = 0 \) and solving \( x^2 + 6x + 9 = 0 \) gives \( x = -3 \). Critical points are \( x = 0 \) and \( x = -3 \).

Step 6: Second Derivative Test for Local Minima and Maxima

Apply the second derivative test at critical points: - For \( x = -3 \), \( f''(-3) = 40(-3)^3 + 180(-3)^2 + 180(-3) < 0 \), so \( f(x) \) has a local maximum at \( x = -3 \).- For \( x = 0 \), \( f''(0) = 180 \), so \( f(x) \) has a local minimum at \( x = 0 \).

Key Concepts

Second Derivative Test

The Second Derivative Test helps us identify whether a critical point is a local minimum or a local maximum of a function. Here's how it works: The second derivative of a function, denoted as \( f''(x) \), provides information about the concavity around the critical points. A critical point is where the first derivative \( f'(x) \) equals zero or is undefined. If \( f''(x) > 0 \) at a critical point, the function is concave up there, indicating a local minimum. Conversely, if \( f''(x) < 0 \), the function is concave down, indicating a local maximum. In our example, we found \( f''(x) = 40x^3 + 180x^2 + 180x \). We need to check the sign of \( f''(x) \) at our critical points from the first derivative. - At \( x = -3 \), \( f''(-3) < 0 \), indicating a local maximum. - At \( x = 0 \), \( f''(0) = 180 > 0 \), indicating a local minimum. By understanding the second derivative's sign, we can determine the nature of critical points more confidently and accurately.

First Derivative

The first derivative of a function \( f(x) \), denoted as \( f'(x) \), is vital for understanding the rate at which \( f(x) \) changes over its domain. We calculate the first derivative to find the slopes of the tangent lines to the curve of \( f(x) \). To obtain \( f'(x) \), we differentiate each term of the function. In our case, for \( f(x) = 2x^5 + 15x^4 + 30x^3 - 42 \), we apply the power rule to each term individually: - Differentiate \( 2x^5 \) to get \( 10x^4 \). - Differentiate \( 15x^4 \) to get \( 60x^3 \). - Differentiate \( 30x^3 \) to get \( 90x^2 \). - The derivative of a constant, \( -42 \), is zero. Thus, \( f'(x) = 10x^4 + 60x^3 + 90x^2 \). By finding the roots of \( f'(x) \), we discover where the function's slope is zero, giving us critical points that help us explore the function's behavior further.

Critical Points

Critical points of a function are where the first derivative \( f'(x) \) is zero or undefined. These points are essential as they indicate where a function may reach a local maximum, local minimum, or change its slope direction dramatically. Critical points are found by solving \( f'(x) = 0 \). In our exercise, we set \( 10x^4 + 60x^3 + 90x^2 = 0 \) and factored it to get \( 10x^2(x^2 + 6x + 9) = 0 \). This results in solutions \( x = 0 \) and \( x = -3 \). These values indicate points on the function \( f(x) \) worth analyzing for maxima, minima, or inflection points. Sometimes additional tools, like the second derivative test, are applied to further understand the behavior at these points.

Concavity Intervals

Concavity intervals describe where a function is bending upwards (concave up) or downwards (concave down) along its graph. The second derivative \( f''(x) \) helps identify these intervals. To find concavity intervals, we solve \( f''(x) = 0 \) and use test points around the roots found to determine the sign of \( f''(x) \). In our solution, \( f''(x) = 40x^3 + 180x^2 + 180x \), which factors to \( 40x(x^2 + 4.5x + 4.5) = 0 \), giving critical points \( x = 0, -3, \) and \( -1.5 \). We tested intervals:

• \( (-\infty, -3) \) where the function is concave down,
• \( (-3, -1.5) \) where it is concave up,
• \( (-1.5, 0) \) where it is concave down,
• \( (0, \infty) \) where it is concave up.

Points of inflection, where concavity changes, occur at these transition points, precisely at \( x = -3, -1.5, \) and \( 0 \). Understanding concavity can help in sketching the curve and comprehending the function's behavior.