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Chapter 4: Problem 45

Follow the outline given in this section to give a careful sketch of the graph of each of the functions in Exercises \(45-54 .\) Your sketch should exhibit, and have labeled, all of the following: (i) all local and global extrema, (ii) inflection points, (iii) intervals on which function is increasing or decreasing, (iv) intervals on which function is concave up or concave down, (v) all horizontal, vertical, and skewasymptotes. $$ f(x)=x^{4}+6 x^{3}+x^{2}+x-4 $$

Short Answer

Expert verified
Graph the polynomial using critical points, inflection points, and sign changes; no asymptotes exist.

Step by step solution

01

Find Critical Points for Extrema

To find the local and global extrema, first find the first derivative of the function to locate critical points. Given, \( f(x) = x^4 + 6x^3 + x^2 + x - 4 \), the first derivative is \( f'(x) = 4x^3 + 18x^2 + 2x + 1 \). Set \( f'(x) = 0 \) to find critical points. Solve the polynomial equation for values of \( x \).
02

Determine Increasing or Decreasing Intervals

Using the critical points from Step 1, test intervals on the number line to determine where \( f(x) \) is increasing or decreasing. Use test points in each interval to evaluate \( f'(x) \). If \( f'(x) > 0 \), the function is increasing; if \( f'(x) < 0 \), it is decreasing.
03

Locate Inflection Points for Concavity

To identify inflection points, find the second derivative, \( f''(x) = 12x^2 + 36x + 2 \). Set \( f''(x) = 0 \) to solve for potential inflection points. Check for sign changes in the second derivative around these points to confirm inflection points, indicating a change from concave up to concave down, or vice versa.
04

Determine Concave Up or Down Intervals

Evaluate the sign of \( f''(x) \) in intervals determined from inflection points. If \( f''(x) > 0 \), the function is concave up in that interval. If \( f''(x) < 0 \), the function is concave down.
05

Identify Asymptotes

Analyze the function to determine if there are any asymptotes. Since \( f(x) \) is a polynomial, it does not have vertical, horizontal, or oblique asymptotes. Polynomials of degree \( n \) behave like \( x^n \) as \( x \to \pm \infty \), where no asymptotes occur.
06

Sketch the Graph

Using all collected information, sketch the graph of \( f(x) \). Mark local maxima and minima from critical points, draw concavity changes at inflection points, and ensure the graph increases or decreases as determined. Label all features accordingly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points in calculus are where the first derivative of a function is zero or undefined. These points help us find the local maximum or minimum points on a graph.
For the function given, the first derivative is \( f'(x) = 4x^3 + 18x^2 + 2x + 1 \).
By setting \( f'(x) = 0 \), we find the values of \( x \) that yield critical points. Evaluating this can be challenging. Often, numeric or graphing tools are used to approximate solutions for polynomial equations of high degree.
Remember, at these critical points, the function can either:
  • Reach a local maximum (a peak point on the graph)
  • Reach a local minimum (a trough point on the graph)
  • Or simply change direction smoothly (neither maximum nor minimum)
Identifying these points is crucial for graph sketching since they tell us where the graph changes its slope direction.
Increasing and Decreasing Intervals
After identifying the critical points, the next task is to determine the intervals where the function is increasing or decreasing.
This involves evaluating the sign of the first derivative, \( f'(x) \).
In particular:
  • If \( f'(x) > 0 \), the function increases on that interval.
  • If \( f'(x) < 0 \), the function decreases.
Choose test points within the intervals defined by the critical points to validate the sign of the derivative. This step is fundamental in sketching because it helps highlight the behavior of the function between critical points. It indicates whether the graph is moving upwards or downwards.
Inflection Points
Inflection points are where a function changes its concavity鈥攆rom concave up to concave down or vice versa.
To find these, you need to compute the second derivative of the given function.
Given \( f(x) = x^4 + 6x^3 + x^2 + x - 4 \), the second derivative is \( f''(x) = 12x^2 + 36x + 2 \).
Setting \( f''(x) = 0 \) allows us to find potential inflection points.
Once those are found, we check if the second derivative changes sign around these points. If it does, an actual inflection point occurs there. Spotting these inflection points is significant for understanding how the graph curves across its domain.
Concavity
Concavity describes whether a graph bends upwards or downwards at any point on a function.
A function is
  • concave up if \( f''(x) > 0 \) at that interval, which often looks like a cup facing upwards,
  • concave down if \( f''(x) < 0 \), resembling an upside-down cup.
By examining intervals set by inflection points, you determine how the graph bends over specified sections.
This analysis contributes to the sketch by ensuring you correctly draw how the curve turns between key points on the graph.
Asymptotes
Asymptotes are lines that a graph approaches but never touches. However, not all functions have them.
For polynomial functions like \( f(x) = x^4 + 6x^3 + x^2 + x - 4 \), no asymptotes exist.
These functions have infinite domains and tend to infinity, without approaching any horizontal, vertical, or skew lines.
It's important to check for asymptotes because they can heavily influence the direction and end behavior of functions. But with polynomials, you can rest easy knowing the graph continues without them.

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Most popular questions from this chapter

Darboux's Theorem (after Jean Darboux, \(1842-1917\) ) states that if \(f^{\prime}\) exists at every point of an open interval containing \([a, b],\) and if \(\gamma\) is between \(f^{\prime}(a)\) and \(f^{\prime}(b),\) then there is a \(c\) in the interval \((a, b)\) such that \(f^{\prime}(c)=\gamma .\) The existence of such a \(c\) follows from the Intermediate Value Theorem if \(f^{\prime}\) is assumed continuous. Darboux's Theorem, which you will prove in this exercise, tells us that the assumption of continuity is not needed for functions that are derivatives. Suppose for definiteness that \(f^{\prime}(a)<\gamma<\) \(f^{\prime}(b) .\) Define \(g\) on \([a, b]\) by \(g(x)=f(x)-\gamma x\) a. Explain why \(g\) has a minimum value that occurs at some point \(c\) in \((a, b)\). b. Show that \(f^{\prime}(c)=\gamma\). c. Use Darboux's Theorem to show that if \(f\) increases on the interval \((l, c)\) and decreases on the interval \((c, r)\) (or vice versa), then \(f^{\prime}(c)=0\)

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