Problem 1 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 1 through 10 , determine intervals of increase and decrease and intervals of concavity for the given function. Then sketch the graph of the function. Be sure to show all key features such as intercepts, asymptotes, high and low points, points of inflection, cusps, and vertical tangents. $$ f(x)=-2 x^{3}+3 x^{2}+12 x-5 $$

Step-by-Step Solution

Answer

f is increasing on \((-1, 2)\), decreasing on \((-\text{infinity}, -1)\) and \((2, \text{infinity})\). It is concave up on \((-\text{infinity}, 1/2)\) and concave down on \((1/2, \text{infinity})\).

Step 1 - Find the first derivative

To determine the intervals of increase and decrease, start by finding the first derivative of the function. Given:\[ f(x) = -2x^3 + 3x^2 + 12x - 5 \]Differentiate with respect to \(x\):\[ f'(x) = \frac{d}{dx}(-2x^3 + 3x^2 + 12x - 5) = -6x^2 + 6x + 12 \]

Step 2 - Find critical points

Set the first derivative equal to zero to find critical points:\[ -6x^2 + 6x + 12 = 0 \]This simplifies to:\[ -6(x^2 - x - 2) = 0 \]Factor the quadratic equation:\[ -6(x - 2)(x + 1) = 0 \]The critical points are \(x = 2\) and \(x = -1\).

Step 3 - Determine intervals of increase and decrease

Use the critical points to determine the intervals of increase and decrease by testing the sign of \(f'(x)\) in each interval: - For \(x \text{ in } (-\text{infinity}, -1)\), choose \(x = -2\), \(f'(-2) = -6(-2^2) + 6(-2) + 12 = -24 - 12 + 12 = -24\) (negative, so decreasing).- For \(x \text{ in } (-1, 2)\), choose \(x = 0\), \(f'(0) = 12\) (positive, so increasing).- For \(x \text{ in } (2, \text{infinity})\), choose \(x = 3\), \(f'(3) = -24\) (negative, so decreasing).

Step 4 - Find the second derivative

To determine concavity, find the second derivative of the function:\[ f''(x) = \frac{d^2}{dx^2}(-2x^3 + 3x^2 + 12x - 5) = -12x + 6 \]

Step 5 - Find points of inflection

Set the second derivative equal to zero to find points of inflection:\[ -12x + 6 = 0 \]Solve for \(x\):\[ x = \frac{1}{2} \]

Step 6 - Determine intervals of concavity

Use the point of inflection to determine the intervals of concavity by testing the sign of \(f''(x)\) in each interval:- For \(x\text{ in } (-\text{infinity}, 1/2)\), choose \(x = 0\), \(f''(0) = 6\) (positive, so concave up).- For \(x \text{ in } (1/2, \text{infinity})\), choose \(x = 1\), \(f''(1) = -6\) (negative, so concave down).

Step 7 - Graph the function

Sketch the graph using the information above:- Intervals of increase: \((-1, 2)\)- Intervals of decrease: \((-\text{infinity}, -1)\) and \((2, \text{infinity})\)- Interval of concave up: \((-\text{infinity}, 1/2)\)- Interval of concave down: \((1/2, \text{infinity})\)- Critical points: \((2, f(2))\) and \((-1, f(-1))\)- Point of inflection: \((1/2, f(1/2))\)- y-intercept: \((0, f(0)) = -5\)Combine all this data to make an accurate sketch.

Key Concepts

Intervals of Increase and Decrease

Intervals of increase and decrease tell us where the function is rising or falling. To find these intervals, you start by finding the first derivative of the function, which is the slope of the tangent at any point. By solving for when this derivative equals zero, you find the critical points. These points help break the function into intervals.

In the given exercise, our function is: f(x) = -2x^3 + 3x^2 + 12x - 5. The first derivative is: f'(x) = -6x^2 + 6x + 12. Setting this equal to zero, we find the critical points: x = 2 and x = -1. Testing points in these intervals, we find:

• From -∞ to -1: the function is decreasing.
• From -1 to 2: the function is increasing.
• From 2 to ∞: the function is decreasing.

This helps us understand how the function behaves in different parts of its domain.

Intervals of Concavity

Intervals of concavity describe whether the graph is curving upwards or downwards. This is found by taking the second derivative of the function. The second derivative tells you about the rate of change of the slope.

For our function, the second derivative is: f''(x) = -12x + 6. Setting this equal to zero, we get the point of inflection: x = 1/2. Checking the sign of the second derivative around this point:

• From -∞ to 1/2: the function is concave up (f''(x) > 0).
• From 1/2 to ∞: the function is concave down (f''(x) < 0).

These intervals of concavity tell us more about the

Critical Points

Critical points are where the first derivative of the function equals zero or is undefined. These points are important because they can be local minima, maxima, or points of inflection.

For our function: f(x) = -2x^3 + 3x^2 + 12x - 5. The first derivative is: f'(x) = -6x^2 + 6x + 12. Setting the first derivative to zero, we get: -6(x-2)(x+1) = 0. Thus, the critical points are x = 2 and x = -1.
These points divide the graph into regions where the function may be increasing or decreasing as we tested in the intervals above. Knowing the exact behavior at these critical points helps sketch the function accurately.

Points of Inflection

Points of inflection are where the function shifts from concave up to concave down or vice versa. Identifying these points requires finding where the second derivative changes sign.

Using our function's second derivative: f''(x) = -12x + 6, and setting it to zero gives us: -12x + 6 = 0 → x = 1/2. This is our point of inflection.

• To the left of 1/2, the graph is concave up.
• To the right of 1/2, the graph is concave down.

Point of inflection provides crucial information about how the shape of the graph changes.

Graph Sketching

Graph sketching involves putting together all the information we've gathered:

• Intervals of increase: (-1, 2)
• Intervals of decrease: (-∞, -1) and (2, ∞)
• Intervals of concave up: (-∞, 1/2)
• Intervals of concave down: (1/2, ∞)
• Critical points: (2, f(2)) and (-1, f(-1))
• Point of inflection: (1/2, f(1/2))
• Y-intercept: (0, f(0)) = (0, -5)

Combine these details carefully to sketch an accurate graph. Pay attention to how the graph curves and changes direction at the critical points and the point of inflection.