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Chapter 10: Problem 39

Sketch the graph of the function, using the curve-sketching quide of this section. $$ h(x)=x^{3}-3 x+1 $$

Short Answer

Expert verified
The function \(h(x) = x^3 - 3x + 1\) has no symmetry, a y-intercept at (0, 1), critical points at x = -1 and x = 1, is increasing on \( (-\infty, -1) \cup (1, \infty) \), decreasing on \((-1, 1)\), and has an inflection point at x = 0. The function is concave up on \( (0, \infty) \) and concave down on \( (-\infty, 0) \). A rough sketch can be drawn by combining this information.

Step by step solution

01

Find Domain and Range

In this case, the function is a polynomial. By definition, the domain of a polynomial function is all real numbers, so the function is defined for every value of x. There are no restrictions on the range, so it can also take any real number as a result. Domain: \( (-\infty, \infty) \) Range: \( (-\infty, \infty) \)
02

Determine Symmetry

To check if the function has any symmetry, we will look for even and odd functions. If the function is even, the graph will be symmetric about the y-axis. If the function is odd, the graph will be symmetric about the origin. \( h(-x) = (-x)^3 - 3(-x) + 1 = -x^3 + 3x + 1 \) The result is not equal to the original function, so it is not an even function. \( -h(x) = -(x^3 - 3x + 1) = -x^3 + 3x - 1 \) The result is also not equal to the original function, so it is not an odd function. Therefore, the function does not have any symmetries.
03

Find Intercepts

To find the x-intercepts, set h(x) equal to 0 and solve for x: $$ 0 = x^3 - 3x + 1 $$ To solve this cubic equation, we can try some integer values and test their possible factors. In this case, there are no integer solutions. So we can't find any simple rational solutions for this equation. It shows that this curve may have an intercept (or two intercepts) with the x-axis, but we won't be able to have a great sketch at this point. Now, to find the y-intercepts, set x equal to 0 and solve for h(x): $$ h(0) = 0^3 - 3(0) + 1 = 1 $$ The y-intercept is (0, 1).
04

Identify Critical Points

To find the critical points, find the first derivative of the function: $$ h'(x) = 3x^2 - 3 $$ Now set the first derivative equal to zero and solve for x: $$ 0 = 3x^2 - 3 $$ Solve for x: $$ x^2 = 1 \\ x = \pm 1 $$ So, there are two critical points at x = -1 and x = 1.
05

Determine Increasing and Decreasing Intervals

Using the critical points, we can find the intervals of increasing and decreasing: For \(x < -1\), \(h'(x) > 0\), so the function is increasing. For \(-1 < x < 1\), \(h'(x) < 0\), so the function is decreasing. For \(x > 1\), \(h'(x) > 0\), so the function is increasing. So the intervals are: Increasing: \( (-\infty, -1) \cup (1, \infty) \) Decreasing: \( (-1, 1) \)
06

Identify Inflection Points and Concavity

Find the second derivative of the function: $$ h''(x) = 6x $$ Now set the second derivative equal to zero and solve for x: $$ 0 = 6x \\ x = 0 $$ There is an inflection point at x = 0. Check the intervals of the second derivative: For \(x < 0\), \(h''(x) < 0\), so the function is concave down. For \(x > 0\), \(h''(x) > 0\), so the function is concave up. So the intervals are: Concave up: \( (0, \infty) \) Concave down: \( (-\infty, 0) \)
07

Sketch the Graph

Now that we have collected all the necessary information, we can sketch the graph: 1. The function has no symmetry. 2. There is a y-intercept at (0, 1). 3. There are critical points at x = -1 and x = 1. 4. The function is increasing on \( (-\infty, -1) \cup (1, \infty) \) and decreasing on \( (-1, 1) \). 5. There is an inflection point at x = 0. The function is concave up on \( (0, \infty) \) and concave down on \( (-\infty, 0) \). By combining all of this information, we can sketch a rough graph of the function \(h(x) = x^3 - 3x + 1 \).

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Key Concepts

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

Polynomial Functions
Polynomial functions are fundamental components of algebra. They are expressions composed of variables and coefficients, involving operations of addition, subtraction, multiplication, and non-negative whole number exponents. The function given in the exercise, \( h(x) = x^3 - 3x + 1 \), is a cubic polynomial. This means its highest degree is three, due to the term \( x^3 \).

Some key characteristics of polynomial functions include:
  • The domain of any polynomial function is always all real numbers \((-\infty, \infty)\) since they are defined for every \(x\).
  • They are continuous and smooth, which means there are no gaps, jumps, or sharp turns.
  • The range can also be all real numbers but depends on the polynomial's behavior.
Understanding these features helps when sketching the graph of polynomial functions, predicting its end behavior (as \(x\) approaches \(+\infty\) or \(-\infty\)), and determining intercepts.
Derivatives
Derivatives are a key tool in calculus for analyzing the behavior of functions. The derivative of a function at any point gives us the slope of the tangent to the curve at that point, which tells us how the function is changing.
  • The first derivative, \( h'(x) \), of a function finds where the function's rate of change (slope) is zero.
  • This point can tell us about critical points, which might be where the function reaches a maximum or minimum, or changes direction.
For the function \( h(x) = x^3 - 3x + 1 \), the first derivative is \( h'(x) = 3x^2 - 3 \). Setting \( h'(x) = 0 \) gives us the critical points.Using derivatives is not just about finding maximums and minimums, but also gives insight into the intervals of the graph where the function is increasing (positive derivative) or decreasing (negative derivative). This information is vital for drawing a detailed graph.
Critical Points
Critical points occur where the derivative equals zero or where it does not exist. These points are important as they may be where the function changes direction, reaches a local maximum, or a local minimum. For our polynomial function, \( x^3 - 3x + 1 \), we set the first derivative \( h'(x) = 3x^2 - 3 \) equal to zero to find critical points.
  • Solving \( 3x^2 - 3 = 0 \) gives us \( x = \pm 1 \).
  • By evaluating these points, we can identify whether these are maximums, minimums, or saddle points.
The process involves studying the changing signs of \( h'(x) \):
  • If \( h'(x) \) changes from positive to negative at a critical point, it's a local maximum.
  • If \( h'(x) \) changes from negative to positive, it's a local minimum.
  • If there's no change, it could be a saddle point.
Exploring these aspects through critical points gives detailed insight into the function's shape and helps in accurately sketching graphs.
Concavity
Concavity informs us about the direction the graph curves and helps conclude where the graph is concave up or down. This is achieved by examining the second derivative of the function.
  • A function is concave up on intervals where its second derivative is positive.
  • Conversely, it is concave down where the second derivative is negative.
  • An inflection point, where the function switches from concave up to down or vice versa, occurs when the second derivative is zero or undefined but changes sign.
For the function \( h(x) = x^3 - 3x + 1 \), the second derivative is \( h''(x) = 6x \). At \( x = 0 \), the second derivative is zero, indicating a potential inflection point, and the concavity changes:
  • For \( x < 0 \), \( h''(x) < 0 \) indicating the graph is concave down.
  • For \( x > 0 \), \( h''(x) > 0 \), signaling the graph is concave up.
Understanding concavity is essential, not only for sketching graphs but also for grasping the subtler aspects of function behavior, such as acceleration and its turning points.

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Most popular questions from this chapter

A stone is thrown straight up from the roof of an \(80-\mathrm{ft}\) building. The height (in feet) of the stone at any time \(t\) (in seconds), measured from the ground, is given by $$ h(t)=-16 t^{2}+64 t+80 $$ What is the maximum height the stone reaches?

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