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Chapter 10: Problem 37

Sketch the graph of the function, using the curve-sketching quide of this section. $$ g(x)=4-3 x-2 x^{3} $$

Short Answer

Expert verified
The graph of the function \(g(x) = 4 - 3x - 2x^3\) has x-intercepts at the solutions of \(4 - 3x - 2x^3 = 0\), a y-intercept at (0, 4), critical points at \(\left(\frac{1}{\sqrt{2}}, g\left(\frac{1}{\sqrt{2}}\right)\right)\) and \(\left(-\frac{1}{\sqrt{2}}, g\left(-\frac{1}{\sqrt{2}}\right)\right)\), and an inflection point at (0, 4). The graph tends towards negative infinity as \(x\) approaches positive infinity and tends towards positive infinity as \(x\) approaches negative infinity. Use this information to sketch the graph, connecting the points smoothly and showing the graph's behavior as x approaches infinity.

Step by step solution

01

Find the x and y-intercepts

To find the x-intercept(s), set \(g(x) = 0\) and solve for \(x\): \[0 = 4 - 3x - 2x^3\] To find the y-intercept, set \(x = 0\) and find the corresponding value of \(g(x)\): \[g(0) = 4 - 3(0) - 2(0)^3 = 4\] So the y-intercept is (0, 4).
02

Find the critical points

To find the critical points, take the first derivative of \(g(x)\) and set it equal to zero: \[g'(x) = -3 - 6x^2\] Set \(g'(x) = 0\): \[-3 - 6x^2 = 0\] Solve for \(x\): \[x^2 = \frac{1}{2}\] \[x = \pm\frac{1}{\sqrt{2}}\] So the critical points are \(\left(\frac{1}{\sqrt{2}}, g\left(\frac{1}{\sqrt{2}}\right)\right)\) and \(\left(-\frac{1}{\sqrt{2}}, g\left(-\frac{1}{\sqrt{2}}\right)\right)\).
03

Find the inflection points

To find the inflection points, take the second derivative of \(g(x)\) and set it equal to zero: \[g''(x) = -12x\] Set \(g''(x) = 0\): \[-12x = 0\] Solve for \(x\): \[x = 0\] The inflection point is \(\left(0, g(0)\right) = (0, 4)\).
04

Determine the graph's behavior as x approaches positive or negative infinity

As \(x\) approaches positive infinity, the term \(-2x^3\) dominates, so the graph will tend towards negative infinity. Likewise, as \(x\) approaches negative infinity, the term \(-2x^3\) dominates, so the graph will tend towards positive infinity.
05

Sketch the graph

Now that we know the x and y-intercepts, critical points, inflection points, and the behavior of the graph as x approaches positive or negative infinity, we can sketch the graph of the function \(g(x) = 4 - 3x - 2x^3\). Plot the points and make sure the graph connects the points smoothly, indicating the critical points and inflection points as necessary, and showing the graph's behavior as x approaches infinity.

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Key Concepts

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

Critical Points
In calculus, when sketching the graph of a function, identifying critical points is important. These are the points on the graph where the slope of the tangent line is zero or where the derivative is undefined.

For our function, let's break down how we found the critical points from the step-by-step solution.

First, we took the derivative of the function:
  • Given function: \(g(x) = 4 - 3x - 2x^3\)
  • First derivative: \(g'(x) = -3 - 6x^2\)
We then set this derivative equal to zero to find the critical points. A quick calculation shows:
  • \(-3 - 6x^2 = 0\)
  • Solving yields \(x = \pm\frac{1}{\sqrt{2}}\)
Once \(x\) values are known, substitute them back into the original function to find corresponding \(y\) values. This gives us the complete critical points, which are points where the graph's slope changes from increasing to decreasing, or vice versa. These points help indicate potential local maxima and minima on your graph.

Critical points are vital because they shape the 'peaks' and 'valleys' of the graph, providing a better understanding of its nature.
Inflection Points
Inflection points are where a graph changes concavity, which means it switches from being "cup-shaped" to "cap-shaped" or the other way round. To find these, you need the second derivative of the function.

In our exercise, this process began by calculating the second derivative:
  • Given the first derivative: \(g'(x) = -3 - 6x^2\)
  • Second derivative: \(g''(x) = -12x\)
Inflection points are found by setting \(g''(x)\) equal to zero:
  • \(-12x = 0\) gives us \(x = 0\)
Substituting \(x = 0\) back into the original function provides the \(y\)-value giving us the inflection point: \((0, 4)\). This single inflection point tells us there's a change in the curvature of the graph at \(x = 0\).

Understanding inflection points allows you to better predict the shape and behavior of a function's graph.
x and y Intercepts
Finding x and y-intercepts is a foundational step in graph sketching. Intercepts are the points where the graph crosses the axes.

The x-intercept occurs where the function equals zero. You are essentially looking for the value of \(x\) that makes the entire function turn into zero. For our function:
  • You set \(g(x) = 0\), resulting in the equation \(0 = 4 - 3x - 2x^3\). Solving this gives \(x\) values where the graph crosses the \(x\)-axis.
The y-intercept is simpler to find:
  • Set \(x = 0\) and solve for \(g(x)\). This gives \(g(0) = 4\). So, the y-intercept is the point \((0, 4)\).
These intercepts are crucial because they provide initial fixed points to plot on the graph, offering a start to understand the overall shape and direction of the graph. Knowing these intercepts helps indicate where the function begins or ends on the given axes, thus contributing to a complete picture of the function's behavior.

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Most popular questions from this chapter

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