Question

Use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. \(f(x)=-0.4 x^{4}-0.5 x^{3}+0.8 x^{2}-2 \quad[-3,2]\)

Short answer

Local minimum at x = 0. Function increases on [-1.85, 0] and [0, 0.55], decreases on [-3, -1.85] and [0.55, 2].

Step-by-step solution

- Determine the derivative

First, find the first derivative of the function, which will help find critical points where local maxima and minima might occur. The function is given as f(x)=-0.4 x^{4}-0.5 x^{3}+0.8 x^{2}-2. Differentiate term by term: f'(x) = -1.6x^3 - 1.5x^2 + 1.6x.

- Find critical points

Set the derivative equal to zero and solve for x to find the critical points: -1.6x^3 - 1.5x^2 + 1.6x = 0. Factor out the common term which is x: x(-1.6x^2 - 1.5x + 1.6) = 0. Solve for x to get the critical points.

- Solve the factored equation

To solve the factored equation x(-1.6x^2 - 1.5x + 1.6) = 0, we get: x = 0, -1.6x^2 - 1.5x + 1.6 = 0. Solve the quadratic equation using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a, where a = -1.6, b = -1.5, c = 1.6.

- Calculate roots of the quadratic

Using the quadratic formula, the roots are: x = (-(-1.5) ± √((-1.5)^2 - 4(-1.6)(1.6))) / (2(-1.6)) x = (1.5 ± √(2.25 + 10.24)) / -3.2 x = (1.5 ± √12.49) / -3.2 Therefore, the additional roots are calculated as x ≈ -1.85 and x ≈ 0.55.

- Determine local maxima and minima using the second derivative test

Evaluate the second derivative, f''(x), to determine concavity at the critical points. If the second derivative at a point is positive, it's a local minimum; if negative, a local maximum. The second derivative is: f''(x) = -4.8x^2 - 3x + 1.6.

- Evaluate the second derivative at critical points

Substitute the critical points (0, -1.85, 0.55) into f''(x): f''(0) = -4.8(0)^2 - 3(0) + 1.6 = 1.6 f''(-1.85) = -4.8(-1.85)^2 - 3(-1.85) + 1.6 ≈ -15.21 f''(0.55) = -4.8(0.55)^2 - 3(0.55) + 1.6 ≈ -2.53 Therefore, (0, f(0)) is a local minimum.

- Use a graphing utility to graph and confirm results

Use a graphing utility to plot the function f(x) = -0.4x^4 - 0.5x^3 + 0.8x^2 - 2 over the interval [-3, 2]. Visually confirm the local maximum and minimum values, as well as intervals where the function is increasing or decreasing.

- Identify Intervals of Increase and Decrease

Using the graph, determine where the function is increasing and where it is decreasing. The function is decreasing on [-3, -1.85] and [0.55, 2], and increasing on [-1.85, 0] and [0, 0.55].

Key concepts

Critical Points

Critical points are where the first derivative of a function is equal to zero or undefined. These points indicate potential local maxima or minima. For our function, the first derivative is found by differentiating term by term:
\[ f'(x) = -1.6x^3 - 1.5x^2 + 1.6x. \]
By setting \( f'(x) = 0 \), we solve:
\(-1.6x^3 - 1.5x^2 + 1.6x = 0.\)
We factor out the common term \( x \):
\[ x(-1.6x^2 - 1.5x + 1.6) = 0. \]
Solving this factored equation, we get the critical points \( x = 0 \), \( x ≈ -1.85 \), and \( x ≈ 0.55 \).

First Derivative

The first derivative \( f'(x) \) helps us determine where the slope of the function is zero, indicating critical points. For the given function, calculating the first derivative:
\[ f'(x) = -1.6x^3 - 1.5x^2 + 1.6x. \]
By evaluating \( f'(x) = 0 \) at the critical points, we identify where potential local maxima or minima can occur. These points help us segment the function for further analysis, checking which intervals are increasing or decreasing.
It’s essential to solve correctly because these points are stepping stones to finding the behavior of the function and solving higher-order derivatives.

Second Derivative Test

The second derivative test uses \( f''(x) \) to determine concavity at the critical points, crucial for evaluating local maxima and minima. For the given function,
first, find the second derivative:
\[ f''(x) = -4.8x^2 - 3x + 1.6. \]
Next, evaluate \( f''(x) \) at our critical points:
- \( f''(0) = 1.6 \) (positive, local minimum)
- \( f''(-1.85) \) ≈ -15.21 (negative, local maximum)
- \( f''(0.55) \) ≈ -2.53 (negative, local maximum)
Thus, determining concavity helps us confirm whether the function curves up or down at these critical points.

Intervals of Increase and Decrease

By analyzing the first derivative \( f'(x) \), we can determine where the function increases or decreases within a given interval. We found that specific critical points divide the function into distinct intervals:

• The function increases where \( f'(x) > 0 \)
• The function decreases where \( f'(x) < 0 \)

From our calculated points:

- Decreasing on \([-3, -1.85]\) and \([0.55, 2]\)
- Increasing on \([-1.85, 0]\) and \([0, 0.55]\)
Graphically verifying this behavior ensures precision, interpreting how the function ascends or descends across specified segments.
Analyzing these intervals provides a complete picture of the function’s behavior.