Problem 8 - 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. $$ f(x)=-2 x^{3}+12 x^{2}+7 $$

Step-by-Step Solution

Answer

Concave up on \((-\infty, 2)\), concave down on \((2, \infty)\), inflection at \(x = 2\), local minimum at \(x = 0\), local maximum at \(x = 4\).

Step 1: Find the First Derivative

To find the critical points of the function, we first need the first derivative. Given the function \( f(x) = -2x^3 + 12x^2 + 7 \), differentiate with respect to \( x \): \[ f'(x) = \frac{d}{dx}(-2x^3 + 12x^2 + 7) = -6x^2 + 24x. \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points: \[ -6x^2 + 24x = 0. \] Factor out \( -6x \): \[ -6x(x - 4) = 0. \] So, the critical points are \( x = 0 \) and \( x = 4 \).

Step 3: Find the Second Derivative

To determine concavity and identify points of inflection, we need the second derivative. Differentiate \( f'(x) = -6x^2 + 24x \): \[ f''(x) = \frac{d}{dx}(-6x^2 + 24x) = -12x + 24. \]

Step 4: Solve for Points of Inflection

Set the second derivative equal to zero to find points of inflection: \[ -12x + 24 = 0 \] which simplifies to \( x = 2 \). This is the potential inflection point.

Step 5: Determine Concave Up/Down Intervals

Test values around \( x = 2 \) in the second derivative to determine concavity:- For \( x < 2 \), choose \( x = 0 \). \( f''(0) = -12(0) + 24 = 24 \), so \( f \) is concave up.- For \( x > 2 \), choose \( x = 3 \). \( f''(3) = -12(3) + 24 = -12 \), so \( f \) is concave down.

Step 6: Apply the Second Derivative Test

Use the second derivative test on critical points:- At \( x = 0 \): \( f''(0) = 24 > 0 \), so \( f(x) \) has a local minimum.- At \( x = 4 \): \( f''(4) = -12(4) + 24 = -24 < 0 \), so \( f(x) \) has a local maximum.

Key Concepts

Concavity Intervals

Understanding concavity is crucial for developing a comprehensive view of a function's behavior. Concavity tells us if a curve bends upwards or downwards over particular intervals. To find concavity intervals, we utilize the second derivative of the function. Starting with the function \(f(x) = -2x^3 + 12x^2 + 7\), its second derivative is \(f''(x) = -12x + 24\).

• If \(f''(x) > 0\), the function is concave up in that interval.
• If \(f''(x) < 0\), the function is concave down.

By substituting values around \(x = 2\) (our point of potential inflection), we found:

• For \(x < 2\), like \(x = 0\), \(f''(0) = 24 > 0\). So, \(f\) is concave up in this region.
• For \(x > 2\), like \(x = 3\), \(f''(3) = -12 < 0\), indicating concave down.

These intervals help us determine where the function bends upward or downward, providing insights into its shape.

Inflection Points

The transition in concavity is marked by points called inflection points. At an inflection point, a function changes from concave up to concave down or vice versa. To find inflection points:

• Set the second derivative to zero: \(-12x + 24 = 0\).
• The potential inflection point is found at \(x = 2\).

To confirm it's an inflection point, check change in sign of the second derivative around \(x = 2\):

• \(f''(x)\) changes from positive to negative around \(x=2\).

Thus, \(x = 2\) is indeed an inflection point. This signals a change in the curve's bending, providing a deeper understanding of the function's dynamic behavior.

Critical Points

Critical points are vital for understanding where a curve changes its slope. They are points where the first derivative of a function equals zero, indicating a potential maximum, minimum, or a saddle point.

• For the function \(f(x) = -2x^3 + 12x^2 + 7\), solve for when first derivative equals zero: \(f'(x) = -6x^2 + 24x = 0\).
• By factoring: \(-6x(x - 4) = 0\), giving \(x = 0\) and \(x = 4\) as critical points.

Critical points are where the function experiences significant changes in direction, making them targets for further analysis using the second derivative test.

Local Extrema

Local extrema are where the function reaches local maximum or minimum values. To identify these points, the second derivative test can be applied to the critical points found.

• At \(x = 0\), where \(f'(0) = 0\), check the second derivative \(f''(0) = 24 > 0\).This indicates a local minimum since the second derivative is positive. The curve is concave up here.
• At \(x = 4\), where \(f'(4) = 0\), check the second derivative \(f''(4) = -24 < 0\).This indicates a local maximum as the second derivative is negative. The curve is concave down at this point.

Understanding local extrema offers keen insights into the highest and lowest points within specified regions, crucial for graphs and real-world scenario applications.