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Chapter 4: Problem 17

a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative. i) \(y=x^{2}-4\) ii) \(y=x^{2}+8 x+15\) iii) \(y=x^{3}-3 x^{2}+4=(x+1)(x-2)^{2}\) iv) \(y=x^{3}-33 x^{2}+216 x=x(x-9)(x-24)\) b. Use Rolle's Theorem to prove that between every two zeros of \(x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\) there lies a zero of $$n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\cdots+a_{1}$$

Short Answer

Expert verified
The zeros of derivatives lie between the zeros of their corresponding polynomials as shown for each given polynomial using Rolle's Theorem.

Step by step solution

01

Find the Zeros of Polynomial i (y=x^2-4)

Factor the polynomial: \(x^2 - 4 = (x - 2)(x + 2)\). The zeros are found by setting each factor equal to zero: \(x - 2 = 0\) gives \(x = 2\), and \(x + 2 = 0\) gives \(x = -2\). The zeros of the polynomial are \(x = -2\) and \(x = 2\).
02

Find the Zeros of Derivative of Polynomial i

Differentiate \(y = x^2 - 4\) to get \(y' = 2x\). Set the derivative to zero to find critical points: \(2x = 0\) gives \(x = 0\). Thus, the zero of the derivative is \(x = 0\).
03

Plot Zeros for Polynomial i

On a number line, place points at \(x = -2\), \(x = 0\), and \(x = 2\) representing the zeros of the original polynomial and its derivative.
04

Find the Zeros of Polynomial ii (y=x^2+8x+15)

Factor the quadratic equation: \(x^2 + 8x + 15 = (x + 3)(x + 5)\). The zeros are found by setting each factor equal to zero: \(x + 3 = 0\) gives \(x = -3\), and \(x + 5 = 0\) gives \(x = -5\). The zeros of the polynomial are \(x = -3\) and \(x = -5\).
05

Find the Zeros of Derivative of Polynomial ii

Differentiate \(y = x^2 + 8x + 15\) to get \(y' = 2x + 8\). Set the derivative to zero: \(2x + 8 = 0\) gives \(x = -4\). Thus, the zero of the derivative is \(x = -4\).
06

Plot Zeros for Polynomial ii

On a number line, place points at \(x = -5\), \(x = -4\), and \(x = -3\) representing the zeros of the original polynomial and its derivative.
07

Find the Zeros of Polynomial iii (y=x^3-3x^2+4)

Factor the expression: \((x + 1)(x - 2)^2 = 0\). Zeros occur at \(x + 1 = 0\), hence \(x = -1\), and \((x - 2)^2 = 0\), hence \(x = 2\). The zeros of the polynomial are \(x = -1\) and \(x = 2\) (multiplicity 2).
08

Find the Zeros of Derivative of Polynomial iii

Differentiate \(y = x^3 - 3x^2 + 4\) to get \(y' = 3x^2 - 6x\). Factor the derivative: \(3x(x - 2)\). Setting it to zero gives zeros at \(x = 0\) and \(x = 2\).
09

Plot Zeros for Polynomial iii

On a number line, place points at \(x = -1\), \(x = 0\), and \(x = 2\) representing zeros, with \(x = 2\) from both the polynomial and its derivative.
10

Find the Zeros of Polynomial iv (y=x^3-33x^2+216x)

Factor the polynomial: \(y = x(x - 9)(x - 24)\). The zeros are found by setting each factor equal to zero: \(x = 0\), \(x = 9\), and \(x = 24\).
11

Find the Zeros of Derivative of Polynomial iv

Differentiate \(y = x^3 - 33x^2 + 216x\) to get \(y' = 3x^2 - 66x + 216\). Solve \(3x^2 - 66x + 216 = 0\) using the quadratic formula: \(x = \frac{33 \pm \sqrt{33^2 - 3 imes 216}}{3}\). This solves to \(x = 6\) and \(x = 12\).
12

Plot Zeros for Polynomial iv

On a number line, place points at \(x = 0\), \(x = 6\), \(x = 9\), \(x = 12\), and \(x = 24\) representing the zeros of the original polynomial and its derivative.
13

Rolle's Theorem Explanation

Rolle's Theorem states that if a function \(f\) is continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a) = f(b)\), then there is at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\). For polynomial equations, this implies that between any two distinct real zeros, there is a zero of its derivative, confirming that the critical points (zeros of derivative) lie between zeros of the polynomial as seen in the steps for each polynomial case.

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Key Concepts

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

Polynomial Zeros
Zeros of a polynomial are those values of the variable where the polynomial evaluates to zero. They are crucial as they help understand the behavior of the polynomial function. In other words, these are the solutions to the polynomial equation. In the equation example, \(y = x^2 - 4\), the zeros are \(x = 2\) and \(x = -2\). Here鈥檚 how to identify them:
  • Factor the polynomial, if possible. For \(y = x^2 - 4\), we factor it as \((x - 2)(x + 2)\).
  • Set each factor equal to zero and solve for \(x\). This gives us the zeros of the polynomial.
By finding these zeros, one can easily plot the points on a number line, giving a clearer picture of where the polynomial touches or crosses the x-axis.
Derivative Zeros
The zeros of the derivative are equally important as they represent critical points where the slope of the function changes. These can be deduced by setting the derivative of the polynomial to zero. For instance, in \(y = x^2 -4\), the derivative is \(y' = 2x\). Setting \(2x = 0\) gives a zero at \(x = 0\). Here are some steps:
  • Find the derivative of the polynomial.
  • Set the derivative expression equal to zero to solve for \(x\).
  • The solutions provide the x-coordinates of the critical points.
These points are significant because around these points the function may exhibit local maxima, minima, or inflection points.
Factoring Polynomials
Factoring is a method to break down polynomials into simpler parts, which when multiplied together give back the original polynomial. It simplifies the process of finding zeros. Consider the polynomial \(y = x^2 + 8x + 15\). It factors into \((x + 3)(x + 5)\). Steps to factor include:
  • Identify two numbers whose product equals the constant term and whose sum equals the coefficient of the linear term.
  • Rewrite the polynomial using these numbers and factor by grouping if necessary.
This method not only assists in identifying zeros but also aids in simplifying polynomial expressions for deeper analysis.
Quadratic Formula
The quadratic formula is a universal method for finding zeros of any quadratic polynomial, given by \(ax^2 + bx + c = 0\). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This tool is especially useful when factoring isn't straightforward. For instance, to find zeros of the polynomial \(y = x^3 - 33x^2 + 216x\), the quadratic part of its derivative \(3x^2 - 66x + 216\) could be solved using the quadratic formula. Here鈥檚 how:
  • Plug values \(a, b,\) and \(c\) from the quadratic into the formula.
  • Calculate the discriminant \(b^2 - 4ac\). If it is positive, there are two real solutions.
  • Solve for \(x\) using the formula.
This method guarantees finding the zeros for any quadratic polynomial, whether easily factorable or not. It鈥檚 a powerful technique for solving polynomial equations, ensuring a complete understanding of its roots.

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Most popular questions from this chapter

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int \frac{1}{2}\left(\csc ^{2} x-\csc x \cot x\right) d x$$

Verify the formulas by differentiation. $$\int \frac{1}{(x+1)^{2}} d x=-\frac{1}{x+1}+C$$

Solve the initial value problems. $$\frac{d y}{d x}=\frac{1}{2 \sqrt{x}}, \quad y(4)=0$$

Sketch the graph of a continuous function \(y=g(x)\) such that a. \(g(2)=2,02,\) and \(g^{\prime}(x) \rightarrow-1^{+}\) as \(x \rightarrow 2^{+}\) b. \(g(2)=2, g^{\prime}<0\) for \(x<2, g^{\prime}(x) \rightarrow-\infty\) as \(x \rightarrow 2^{-}\) \(g^{\prime}>0\) for \(x>2,\) and \(g^{\prime}(x) \rightarrow \infty\) as \(x \rightarrow 2^{+}\)

Finding displacement from an antiderivative of velocity a. Suppose that the velocity of a body moving along the \(s\) -axis is $$\frac{d s}{d t}=v=9.8 t-3$$ i) Find the body's displacement over the time interval from \(t=1\) to \(t=3\) given that \(s=5\) when \(t=0\). ii) Find the body's displacement from \(t=1\) to \(t=3\) given that \(s=-2\) when \(t=0\). iii) Now find the body's displacement from \(t=1\) to \(t=3\) given that \(s=s_{0}\) when \(t=0\). b. Suppose that the position \(s\) of a body moving along a coordinate line is a differentiable function of time \(t .\) Is it true that once you know an antiderivative of the velocity function ds/dt, you can find the body's displacement from \(t=a\) to \(t=b\) even if you do not know the body's exact position at either of those times? Give reasons for your answer.
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