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Chapter 12: Problem 12

Determine whether the given value is a zero of the function. $$f(t)=1+2 t+t^{3}-t^{5} ; t=\sqrt{2}$$

Short Answer

Expert verified
The value \( t = \sqrt{2} \) is not a zero of the function.

Step by step solution

01

Substitute the Value

Start by substituting \( t = \sqrt{2} \) into the function \( f(t) = 1 + 2t + t^3 - t^5 \). This will allow us to see if the function equals zero for this value.
02

Simplify Each Term

Calculate the value of each term separately:\(\begin{align*}2t & = 2(\sqrt{2}) = 2\sqrt{2}, \t^3 & = (\sqrt{2})^3 = 2\sqrt{2}, \t^5 & = (\sqrt{2})^5 = 4\sqrt{2}.\end{align*}\)
03

Substitute Terms Back

Substitute these values back into the function:\[f(\sqrt{2}) = 1 + 2\sqrt{2} + 2\sqrt{2} - 4\sqrt{2}.\]
04

Combine Like Terms

Combine like terms in the equation:\[1 + 2\sqrt{2} + 2\sqrt{2} - 4\sqrt{2} = 1 + 0 = 1.\]
05

Conclusion

Since \( f(\sqrt{2}) = 1 \), the value \( t = \sqrt{2} \) is not a zero of the function.

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Key Concepts

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

Polynomials
A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. In simpler terms, it can be thought of as a string of numbers and variables combined through addition, subtraction, and multiplication. For example, the given function \( f(t) = 1 + 2t + t^3 - t^5 \) is a polynomial with variable \( t \). The terms are ordered by their degree, which is the power of \( t \):
  • Constant term: \( 1 \), which can also be considered as \( t^0 \).
  • Linear term: \( 2t \) where \( t \) is raised to the power of 1.
  • Cubic term: \( t^3 \), as \( t \) is raised to the power of 3.
  • Quintic term: \( -t^5 \), where \( t \) is raised to the power of 5, the highest degree in this polynomial.
Polynomials serve as a foundation in algebra, making it crucial to understand their structure and behavior, especially when dealing with functions and finding their zeros.
Function Evaluation
Function evaluation is the process of computing the output of a function given an input. It involves substituting the input value into the function and simplifying to find the result. Let's see how it applies to our example function \( f(t) = 1 + 2t + t^3 - t^5 \) with a specific value \( t = \sqrt{2} \).By substituting \( t = \sqrt{2} \) into the function, each occurrence of \( t \) is replaced:
  • Evaluate \( 2t \): Multiply 2 by \( \sqrt{2} \) to get \( 2\sqrt{2} \).
  • Evaluate \( t^3 \): Compute \( (\sqrt{2})^3 = 2\sqrt{2} \).
  • Evaluate \( t^5 \): Calculate \( (\sqrt{2})^5 = 4\sqrt{2} \).
After evaluating each term, substitute the values back to see if the polynomial equals zero. In this case, it does not, as the sum equals 1, not 0.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying expressions to make calculations easier or to derive new expressions. In the context of polynomials and zeroes of functions, it frequently involves combining like terms. In our problem, after substituting \( t = \sqrt{2} \) into the polynomial, we simplified the expression \( 1 + 2\sqrt{2} + 2\sqrt{2} - 4\sqrt{2} \). Here's how algebraic manipulation works here:
  • Identify and group similar terms, in this case, those involving \( \sqrt{2} \).
  • Add and subtract the coefficients of these like terms: \( 2\sqrt{2} + 2\sqrt{2} - 4\sqrt{2} = 0 \).
  • Combine the like terms with the constant: \( 1 + 0 = 1 \).
The final outcome shows that the expression simplifies to 1, confirming \( \sqrt{2} \) is not a zero. Mastery of algebraic manipulation is vital for evaluating functions and understanding their properties.

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Most popular questions from this chapter

You are given an improper rational expression. First, use long division to rewrite the expression in the form (polynomial) \(+\) (proper rational expression) Next, obtain the partial fraction decomposition for the proper rational expression. Finally, rewrite the given improper rational expression in the form (polynomial) \(+\) (partial fractions) $$\frac{2 x^{5}-11 x^{4}-4 x^{3}+53 x^{2}-24 x-5}{2 x^{3}+x^{2}-10 x-5}$$

Find all roots of each equation. Hints: First, factor by grouping. In Exercises 71 and 72 each equation has three roots; in Exercise 73 the equation has six roots; in Exercise 74 there are five roots. $$x^{5}+4 x^{3}+8 x^{2}+32=0$$

One root of the equation \(x^{2}+b x+1=0\) is twice the other; find \(b .\) (There are two answers.)

Scipio Ferro of Bologna well-nigh thirty years ago discovered this rule and handed it on to Antonio Maria Fior of Venice, whose contest with Niccol貌 Tartaglia of Brescia gave Niccol貌 occasion to discover it. He / Tartaglial gave it to me in response to my entreaties, though withholding the demonstration. Armed with this assistance, I sought out its demonstration in /various / forms. - Girolamo Cardano, Ars Magna (Nuremberg, 1545 ) The quotation is from the translation of Ars Magna by T. Richard Witmer (New York: Dover Publications, 1993 ). In his book Ars Magna (The Great Art) the Renaissance mathematician Girolamo Cardano \((1501-1576)\) gave the following formula for a root of the equation \(x^{3}+a x=b\). $$x=\sqrt[3]{\frac{b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}-\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^{2}}{4}+\frac{a^{3}}{27}}}$$ (a) Use this formula and your calculator to compute a root of the cubic equation \(x^{3}+3 x=76\) (b) Use a graph to check the answer in part (a). That is, graph the function \(y=x^{3}+3 x-76,\) and note the \(x-\) intercept. Also check the answer simply by substituting it in the equation \(x^{3}+3 x=76\)

\(\frac{x^{3}+4 x^{2}+16 x}{x^{4}+16}\) Hint: To factor the denominator, add and subtract the term \(16 x^{2}\)
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