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Show that the real zeros of each polynomial function satisfy the given conditions. \(f(x)=x^{5}+2 x^{3}-2 x^{2}+5 x+5 ;\) no real zero less than \(-1\)

Short Answer

Expert verified
There are no real zeros of \( f(x) \) less than \(-1\).

Step by step solution

01

Identify the polynomial function

The given polynomial function is: \[ f(x) = x^5 + 2x^3 - 2x^2 + 5x + 5 \]
02

Analyze the condition

We need to show that there are no real zeros of the polynomial less than \(-1\). This means \(f(x)\) does not cross the x-axis for \(x < -1\).
03

Evaluate the polynomial at test points less than \(-1\)

Let's test some values for \(x \) less than \(-1\) to check if \(f(x)\) can be zero. Starting with \(x = -2\): \[ f(-2) = (-2)^5 + 2(-2)^3 - 2(-2)^2 + 5(-2) + 5 \] \[ = -32 - 16 - 8 - 10 + 5 \] \[ = -61 \] Since \(f(-2) < 0 \), it is not a zero.
04

Evaluate another point

Now trying \(x = -1\): \[ f(-1) = (-1)^5 + 2(-1)^3 - 2(-1)^2 + 5(-1) + 5 \] \[ = -1 - 2 - 2 - 5 + 5 \] \[ = -5 \] This value is also not zero and thus not a real zero.
05

Apply the Intermediate Value Theorem (IVT)

For any continuous function \(f(x)\), the IVT states that if \(f(a)\) and \(f(b)\) have opposite signs, then there is at least one real zero between \(a\) and \(b\). Since \[ f(-2) = -61 \] and \[ f(-1) = -5 \] are both negative, no real zeros exist between \(-2\) and \(-1\) or for any values less than \(-1\).
06

Conclusion

Thus, there are no real zeros of \( f(x) = x^5 + 2x^3 - 2x^2 + 5x + 5 \) that are less than \(-1\).

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Key Concepts

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

polynomial function analysis
Polynomial function analysis helps you understand how the function behaves across different values of x. To analyze a polynomial, you often start by looking at its degree and coefficients. For the function we are working with, f(x) = x^5 + 2x^3 - 2x^2 + 5x + 5, the highest power of x is 5, making it a polynomial of degree 5. This means the graph of the function might have up to 5 real roots and can change direction up to four times due to the x-to-the-power terms.
To find the real zeros of the polynomial, you calculate where the function's value is zero, meaning where the function crosses the x-axis. By testing different values of x or by using methods such as synthetic division or graphing, you can identify these zeros. This analysis, step-by-step, helps in understanding how the polynomial graph behaves overall.
Intermediate Value Theorem
The Intermediate Value Theorem (IVT) is a critical tool in finding if and where zeros of a continuous polynomial function exist within a given interval. The IVT states that if a continuous function, f(x), takes on different signs at two points, a and b (i.e., f(a) and f(b) have opposite signs), there must be at least one point c between a and b such that f(c) = 0.
To apply the IVT to our problem, we checked f(x) at different points. We found that f(-2) = -61 and f(-1) = -5, both negative values, indicating no sign change between -2 and -1. Because both values are negative, the theorem tells us that no zero exists between these points. Similarly, evaluating more negative x values (-3, -4, etc.) can help confirm no sign change.
This confirmed our function f(x) does not have real zeros less than -1, given no sign change between the tested points, concluding no zero in that range.
zero of a function
A zero of a function is an x-value where the function's graph touches or crosses the x-axis. For polynomials, it visually represents where f(x) = 0. Zeros can be found by setting the polynomial equal to zero and solving for x.
In our problem, we tested different x-values to check for zeros. By calculating f(x) at x = -2 and x = -1, we determined that neither value results in f(x) = 0. Evaluating further showed that all function values less than -1 fall below zero. By thoroughly analyzing these points, we deduced the polynomial does not have any real zeros less than -1.
Understanding the zero points helps in graphing the polynomial and solving real-world problems where these points indicate clear critical conditions, such as where a physical object's trajectory hits ground level or when profit/loss reaches break-even.

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Most popular questions from this chapter

The remainder theorem indicates that when a polynomial \(f(x)\) is divided by \(x-k\) the remainder is equal to \(f(k) .\) For $$f(x)=x^{3}-2 x^{2}-x+2$$ use the remainder theorem to find each of the following. Then determine the coordinates of the corresponding point on the graph of \(f(x)\) $$f(2)$$

Find all complex zeros of each polynomial function. Give exact values. List multiple zeros as necessary. $$f(x)=x^{3}-14 x+8$$

Solve each problem. AIDS Cases in the United States The table* lists the total (cumulative) number of AIDS cases diagnosed in the United States through \(2007 .\) For example, a total of \(361,509\) AIDS cases were diagnosed through 1993 (a) Plot the data. Let \(x=0\) correspond to the year 1990 . (b) Would a linear or a quadratic function model the data better? Explain. (c) Find a quadratic function defined by \(f(x)=a x^{2}+b x+c\) that models the data. (d) Plot the data together with \(f\) on the same coordinate plane. How well does \(f\) model the number of AIDS cases? (e) Use \(f\) to estimate the total number of AIDS cases diagnosed in the years 2009 and 2010 (f) According to the model, how many new cases were diagnosed in the year \(2010 ?\) $$\begin{array}{c|c||c|c} \text { Year } & \text { AIDS Cases } & \text { Year } & \text { AIDS Cases } \\\ \hline 1990 & 193,245 & 1999 & 718,676 \\ \hline 1991 & 248,023 & 2000 & 759,434 \\ \hline 1992 & 315,329 & 2001 & 801,302 \\ \hline 1993 & 361,509 & 2002 & 844,047 \\ \hline 1994 & 441,406 & 2003 & 888,279 \\ \hline 1995 & 515,586 & 2004 & 932,387 \\ \hline 1996 & 584,394 & 2005 & 978,056 \\ \hline 1997 & 632,249 & 2006 & 982,498 \\ \hline 1998 & 673,572 & 2007 & 1,018,428 \\ \hline \end{array}$$

Match the rational function in Column I with the appropriate description in Column II. Choices in Column II can be used only once. A. The \(x\) -intercept is \(-3\) B. The \(y\) -intercept is 5 C. The horizontal asymptote is \(y=4\) D. The vertical asymptote is \(x=-1\) E. There is a "hole" in its graph at \(x=-4\) F. The graph has an oblique asymptote. G. The \(x\) -axis is its horizontal asymptote, and the \(y\) -axis is not its vertical asymptote. H. The \(x\) -axis is its horizontal asymptote, and the \(y\) -axis is its vertical asymptote. $$f(x)=\frac{x+3}{x-6}$$

Solve each problem. Height of an Object If an object is projected upward from ground level with an initial velocity of 32 ft per sec, then its height in feet after \(t\) seconds is given by $$ s(t)=-16 t^{2}+32 t $$ Find the number of seconds it will take to reach its maximum height. What is this maximum height?

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