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Chapter 3: Problem 27

Use the Intermediate Value Theorem to approximate the zero of \(f\) in the interval \([a, b]\). Give your approximation to the nearest tenth. (If you have a graphing utility, use it to help you approximate the zero.) $$f(x)=x^{3}+x-1, \quad[0,1]$$

Short Answer

Expert verified
Approximately, the zero of the function \(f(x) = x^{3} + x - 1\) in the interval \([0,1]\) is 0.6.

Step by step solution

01

Evaluating the Function at Interval Ends

Evaluate the function at \(a = 0\) and \(b = 1\) to get \(f(0)\) and \(f(1)\). This means substitute \(0\) and \(1\) into the given function \(f(x)\). Calculate \(f(0)\) by setting \(x = 0\) in \(f(x)\) to get \(0^{3} + 0 - 1 = -1\). Similarly, find \(f(1)\) by setting \(x = 1\) in \(f(x)\) to get \(1^{3} + 1 - 1 = 1\). Since \(f(a)\) and \(f(b)\) encompass \(0\) (as \(f(a) < 0\) and \(f(b) > 0\) or vice versa), we can confirm that the function must cross the \(x\)-axis in the interval \([a, b] = [0, 1]\)
02

Using the Intermediate Value Theorem

Now the zero of \(f\) on the interval \([a, b]\) must be approximated. Check the midpoint of the interval, which is \((0+1)/2 = 0.5 \) and evaluate \(f\) at this new value. Substituting \(x = 0.5\) into the function \(f(x)\), we get \(f(0.5) = (0.5)^{3} + 0.5 - 1 = -0.375\). This is less than \(0\), which indicates that the zero of \(f\) still lies somewhere between \(0.5\) and \(1\). Narrow down the interval to \([0.5, 1]\) and repeat the above process till the required decimal accuracy is achieved.
03

Repeating the Process to Approximate Zero

Now, evaluate the function at the new midpoint, which is \(0.75\) (average of \(0.5\) and \(1\)). When substituting \(x = 0.75\) into \(f(x)\), you obtain \(f(0.75) = (0.75)^{3} + 0.75 - 1 = 0.171875\), which is greater than zero. Thus, now you know the zero lies between \(0.5\) and \(0.75\). Next, take the midpoint between \(0.5\) and \(0.75\) which is \(0.625\) and evaluate \(f\) at \(0.625\) to determine the exact interval where the zero lies. Repeat this process until finding an approximation of the zero to the nearest tenth is achieved.

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Key Concepts

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

Zero of a Function
A zero of a function, also known as a root, is a value of the variable that makes the function equal to zero. In other words, when you substitute this value into the function, the result is zero. Finding zeros is crucial in understanding the behavior of functions, especially when dealing with their graphs. Zeros tell us where the graph of the function will intersect the x-axis.

For polynomial functions like the one in this exercise, zeros can also give insight into factors of the polynomial. Remember, a zero of a function tells us something important about where the output changes sign. It's where the function is neither positive nor negative.
  • Zeros are often found using solving techniques like factoring, graphing, or calculus-based methods like the Newton-Raphson method.
  • The Intermediate Value Theorem is particularly valuable for finding zeros in continuous functions over a specific interval.
Polynomial Functions
Polynomial functions are equations that involve powers of a variable, called "polynomials." These equations form the backbone of algebra and are characterized by the sum of several terms that have the same variable raised to different powers. Typical polynomial functions could be linear, quadratic, cubic, or even higher degrees.

Here's what you need to know about polynomial functions:
  • The degree of the polynomial, indicated by the highest power of the variable, dictates the number of roots or zeros and the general shape of its graph.
  • Cubic polynomials, such as the given function in this exercise, can have up to three real roots.
  • The coefficients in front of each term help determine the polynomial's behavior as the variable approaches positive or negative infinity.
In the exercise, the polynomial function is of degree three, hence why it's called a cubic function. These functions can appear wavy on a graph due to their bending nature around their zeros.
Approximating Zeros
Approximating zeros involves finding a value where the function gets close to zero but might not be precisely zero, due to the constraints like lack of exact factors. In context, the Intermediate Value Theorem is employed to offer a systematic approach to zero approximation by checking function values at specific points.

Here's a simplified way to approximate zeros:
  • Select an interval where the function changes sign, meaning at one end of the interval the function value is positive and at the other end it's negative.
  • Calculate the function at the midpoint of this interval. This is your first approximation step.
  • If the function value at the midpoint is close enough to zero (within a desired accuracy), you've found your approximate zero.
  • If not, adjust your interval based on the sign of the function at the midpoint, and repeat the process, halving the interval each time.
The repeated halving of the interval, as shown in the solution, hones in on the zero. By gradually narrowing down the interval and checking at each midpoint, an approximation to any decimal precision can be achieved.

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Most popular questions from this chapter

Analyzing a Graph In Exercises \(47-58\), analyze the graph of the function algebraically and use the results to sketch the graph by hand. Then use a graphing utility to confirm your sketch. $$f(x)=x(x-2)^{2}(x+1)$$

Credit Cards The numbers of active American Express cards \(C\) (in millions) in the years 1997 to 2006 are shown in the table. (Sourze: American Express) $$ \begin{aligned} &\begin{array}{|l|l|l|l|l|l|} \hline \text { Year } & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \text { Cards, C } & 42.7 & 42.7 & 46.0 & 51.7 & 55.2 \\ \hline \end{array}\\\ &\begin{array}{|l|l|l|l|l|l|} \hline \text { Year } & 2002 & 2003 & 2004 & 2005 & 2006 \\ \hline \text { Cards, C } & 57.3 & 60.5 & 65.4 & 71.0 & 78.0 \\ \hline \end{array} \end{aligned} $$ (a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=7\) corresponding to \(1997 .\) (b) Use what you know about end behavior and the scatter plot from part (a) to predict the sign of the leading coefficient of a quartic model for \(C\). (c) Use the regression feature of a graphing utility to find a quartic model for \(C\). Does your model agree with your answer from part (b)? (d) Use a graphing utility to graph the model from part (c). Use the graph to predict the year in which the number of active American Express cards would be about 92 million. Is your prediction reasonable?

Regression Problem Let \(x\) be the number of units (in tens of thousands) that a computer company produces and let \(p(x)\) be the profit (in hundreds of thousands of dollars). The table shows the profits for different levels of production. $$ \begin{aligned} &\begin{array}{|l|l|l|l|l|l|} \hline \text { Units, } x & 2 & 4 & 6 & 8 & 10 \\ \hline \text { Profit, } p(x) & 270.5 & 307.8 & 320.1 & 329.2 & 325.0 \\ \hline \end{array}\\\ &\begin{array}{|l|l|l|l|l|l|} \hline \text { Units, } x & 12 & 14 & 16 & 18 & 20 \\ \hline \text { Profit, } p(x) & 311.2 & 287.8 & 254.8 & 212.2 & 160.0 \\ \hline \end{array} \end{aligned} $$ (a) Use a graphing utility to create a scatter plot of the data. (b) Use the regression feature of a graphing utility to find a quadratic model for \(p(x)\). (c) Use a graphing utility to graph your model for \(p(x)\) with the scatter plot of the data. (d) Find the vertex of the graph of the model from part (c). Interpret its meaning in the context of the problem. (e) With these data and this model, the profit begins to decrease. Discuss how it is possible for production to increase and profit to decrease.

Analyzing a Graph In Exercises \(47-58\), analyze the graph of the function algebraically and use the results to sketch the graph by hand. Then use a graphing utility to confirm your sketch. $$f(x)=1-x^{6}$$

Determine (a) the maximum number of turning points of the graph of the function and (b) the maximum number of real zeros of the function. $$f(x)=-3 x^{4}+1$$
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