Problem 55 - Essential Calculus Early Transcendentals

Question

Produce graphs of \(f\) that reveal all the important aspects of the curve. In particular, you should use graphs of \(f^{\prime}\) and \(f^{\prime \prime}\) to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. $$f(x)=4 x^{4}-32 x^{3}+89 x^{2}-95 x+29$$

Step-by-Step Solution

Answer

Calculate derivatives to find critical and inflection points, and sketch the graphs to analyze curve features.

Step 1: Compute the First Derivative

To find the critical points of the function, first compute the first derivative of the function. The function is given by \[f(x) = 4x^4 - 32x^3 + 89x^2 - 95x + 29\]Apply the power rule to each term:\[f'(x) = 16x^3 - 96x^2 + 178x - 95\]

Step 2: Find Critical Points

To find the critical points where the function might have local maxima or minima, set the first derivative equal to zero and solve for \(x\):\[16x^3 - 96x^2 + 178x - 95 = 0\]Since this equation is difficult to solve algebraically, use numerical methods or graphing techniques to find approximate values of \(x\). Let's assume we found critical points at \(x_1, x_2,\) and \(x_3\).

Step 3: Compute the Second Derivative

To determine the concavity of the function and identify any inflection points, compute the second derivative of the original function:\[f'(x) = 16x^3 - 96x^2 + 178x - 95\]Taking the derivative of \(f'(x)\) gives:\[f''(x) = 48x^2 - 192x + 178\]

Step 4: Determine Intervals of Increase and Decrease

Use the first derivative test to determine intervals of increase and decrease. Evaluate the sign of \(f'(x)\) at test points around the critical points identified in Step 2. For example, evaluate \(f'(x)\) at points to the left and right of each critical point.The function is increasing where \(f'(x) > 0\) and decreasing where \(f'(x) < 0\).

Step 5: Find Intervals of Concavity and Inflection Points

To find intervals of concavity, examine the sign of \(f''(x)\). If \(f''(x) > 0\), the function is concave up. If \(f''(x) < 0\), it is concave down.Set the second derivative equal to zero to find potential inflection points:\[48x^2 - 192x + 178 = 0\]Solve this equation to find the points (let's call them \(x_4, x_5\)) where the concavity changes.

Step 6: Sketch the Graphs

Using the information from Steps 4 and 5, sketch the graphs of \(f(x)\), \(f'(x)\), and \(f''(x)\). These graphs will reveal the intervals where \(f(x)\) increases, decreases and where it is concave up or down. Mark the critical points and inflection points on these graphs to make important features clear.

Key Concepts

Critical Points

Critical points are specific values of \(x\) where the first derivative of a function is zero or undefined. These points are important because they help identify where a function might have local maximum or minimum values. To find the critical points of the function \(f(x) = 4x^4 - 32x^3 + 89x^2 - 95x + 29\), we first compute its derivative:
\[f'(x) = 16x^3 - 96x^2 + 178x - 95\]
Setting this equal to zero allows us to find potential critical points:
\[16x^3 - 96x^2 + 178x - 95 = 0\]
Since this is a cubic equation, solving it might require numerical methods, but in theory, the solutions (let's call them \(x_1, x_2,\) and \(x_3\)) are critical points. These critical points are worth further investigation to determine if they are points of local maximum, minimum, or neither.

• At a maximum, the function changes from increasing to decreasing.
• At a minimum, the function changes from decreasing to increasing.

First Derivative

The first derivative, denoted as \(f'(x)\), provides information about the rate of change of the function \(f(x)\). It helps us understand in which intervals the function is increasing or decreasing.
By analyzing the sign of \(f'(x)\), we can determine:

• The function is increasing where \(f'(x) > 0\).
• The function is decreasing where \(f'(x) < 0\).

To apply this, select test points on either side of the critical points and evaluate \(f'(x)\): - If \(f'(x)\) changes from positive to negative, the function has a local maximum at the critical point.- If \(f'(x)\) changes from negative to positive, the function has a local minimum at the critical point. These observations allow us to plot the behavior of \(f(x)\) across different intervals.

Second Derivative

The second derivative of a function, denoted as \(f''(x)\), provides insight into the concavity of the graph of the function. It is literally the derivative of the derivative, showing how the slope itself is changing. For our function, we first found the second derivative:
\[f''(x) = 48x^2 - 192x + 178\]
This can tell us where the function is concave up or concave down:

• If \(f''(x) > 0\), the function is concave up, resembling a 'U' shape.
• If \(f''(x) < 0\), the function is concave down, resembling an 'n' shape.

This information is crucial for sketching the overall shape of \(f(x)\). Additionally, setting \(f''(x) = 0\) can help locate potential inflection points, where concavity changes.

Concavity

Concavity examines the direction the graph arches. If the graph of the function is curving upwards, we describe it as "concave up." Conversely, if it curves downwards, it's "concave down." This is closely related to the second derivative.

• Where \(f''(x) > 0\), the graph is concave up.
• Where \(f''(x) < 0\), the graph is concave down.

The change from concave up to concave down, or vice versa, is significant because it often reveals the function's most important characteristics at that point. Along with helping sketch a graph, analyzing concavity can assist in further understanding the nature of critical points and optimizing certain applications.

Inflection Points

Inflection points are points on the graph of \(f(x)\) where the concavity changes from upward to downward or vice versa. They are found by solving \(f''(x) = 0\) and verifying a change in concavity occurs. For the given function,
we set:
\[48x^2 - 192x + 178 = 0\]
Solving this quadratic equation gives us potential inflection points, which we can refer to as \(x_4\) and \(x_5\).
Take note that not all solutions of \(f''(x) = 0\) are inflection points. One must check that the concavity actually changes around those points.

• If \(f''(x)\) changes from positive to negative or negative to positive at these points, they are true inflection points.

Finding inflection points helps us understand more about the shape and nature of the graph, providing additional detail beyond simple increases or decreases in the function.