Transcript of "08 functions and their graphs - part 3"
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Polynomial Functions and Equations<br />Continuation<br />
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Polynomial Functions<br />A polynomial function is a function of the form:<br />n must be a positive integer<br />an ≠ 0<br />All of these coefficients are real numbers<br />The degree of the polynomial is the largest power on any x term in the polynomial.<br />
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Given f(x) =an xn + an-1 xn-1+…+a1 x +a0, <br /> where an ≠ 0 and r is a real number, <br />the following statements are equivalent:<br /><ul><li>x = r is a zero of the function f.
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x = r is a solution of the polynomial equation f(x) = 0.
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(r , 0) is an x-interceptof the graph of f. </li></li></ul><li>Rational zeros of polynomial functions<br />
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Rational Zeros of Polynomial Functions<br />The zeros of a polynomial function are the solutions that can be found when each of the factors of the polynomial is set equal to zero and the value of the variable is solved.<br />
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Fundamental Theorem of Algebra<br />Every polynomial function whose defining equation is f(x) =an xn + an-1 xn-1+…+ a1 x + a0, where an ≠ 0 and n ≥ 1 has at least one complex zero.<br />
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Fundamental Theorem of Algebra<br />Every polynomial function whose defining equation is f(x) =an xn + an-1 xn-1+…+ a1 x + a0, where an ≠ 0 and n ≥ 1 has at least one complex zero.<br />Note that a complex zero of the polynomial can either be a real or imaginary number.<br />
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Rational Zeros of Polynomial Functions<br />The factored form of the polynomial<br /> f(x) =an xn + an-1 xn-1+…+ a1 x +a0<br /> is<br /> f(x) =an (x – r1) (x – r2)… (x – rn)<br /> where r1, r2, … rnare the zeros of the polynomial. <br />Ifa factor (x – r) occurs ktimes, <br />then ris called a zero of multiplicity k.<br />
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Example 1a.<br />Find the zeros of the given function and determine the multiplicity of each zero.<br />Solution:<br />-1 is a zero of multiplicity 3 and <br />1 is a zero of multiplicity 1.<br />
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Example 1b.<br />Find the zeros of the given function and determine the multiplicity of each zero.<br />Solution:<br />1 is a zero of multiplicity 2, <br />-3 is a zero of multiplicity 3, and <br />2 is a zero of multiplicity 1.<br />
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Example 1c.<br />Find the zeros of the given function and determine the multiplicity of each zero.<br />Solution:<br />½ and -½ are zeros of multiplicity 1, <br />3/2 is a zero of multiplicity 4, and <br />-3 is a zero with multiplicity 5.<br />
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Example 2a.<br />Find a polynomial equation of least possible degree which satisfies the given condition: <br />zeros are 2, 1, and ¼<br />Solution:<br />If f(r ) = 0, then (x – r) is a factor of f(x). <br />Therefore, the desired polynomial is <br />f(x) = (x – 2)(x – 1)( x – ¼).<br />
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Example 2b.<br />Find a polynomial equation of least possible degree which satisfies the given condition: <br />zeros are -1, 3 and 1 of multiplicity 2<br />Solution:<br />If f(r ) = 0, then (x – r) is a factor of f(x). <br />Therefore, the desired polynomial is <br />f(x) = (x +1)(x – 3)(x – 1)2<br />
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Example 2c.<br />Find a polynomial equation of least possible degree which satisfies the given condition: <br />f(1) = f(2) = f(3) = f(0) = 0, f(4) = 24<br />Solution:<br />By the Factor Theorem, f(x) = an (x – r1) (x – r2)… (x – rn)<br />Since f(1) = f(2) = f(3) = f(0) = 0, then (x – 0), <br /> (x – 1), (x – 2), and (x – 3) are factors of the polynomial f(x). <br />If f(4) = 24, then<br /> 24 = an (4 – 0)(4 – 1)(4 – 2)(4 – 3) <br />Therefore, the desired polynomial is <br />f(x) = x(x – 1)(x – 2)(x – 3) <br />
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Descartes’s Rule of Signs<br />If f(x) =an xn + an-1 xn-1+…+ a1 x +a0 is a polynomial with real number coefficients, then the following holds true:<br />the number of positive real zeros of f iseither equal to the number of sign changes of f (x)or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero.<br /> the number of negative real zeros of f iseither equal to the number of sign changes of f (-x) or is less than that number by an even integer. If f (-x)has only one variation in sign, then f has exactly one negative real zero. <br />
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Example 3a.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 3rd , there are 3 zeros of the function.<br />There is no variation of signs in f(x) so there no positive roots.<br />This implies that all roots are negative.<br />Now, <br />There are 3 variations of signs in f(x) so there are 3 negative roots.<br />Therefore, all 3 roots of the function are negative.<br />
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Example 3b.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 3rd degree, there are 3 zeros of the function.<br />There is only 1 variation of signs in f(x) so there are either 1 positive root or none at all.<br />Now, <br />There are 2 variations of signs in f(x) so there are either 2 negative roots or none at all.<br />Therefore, there are 1 positive and 2 negative roots of the function.<br />
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Example 3c.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 4th degree, there are 4 zeros of the function.<br />There are 2 variations of signs in f(x) so there are either 2 positive roots or none at all.<br />Now, <br />There are 2 variations of signs in f(x) so there are either 2 negative roots or none at all.<br />Therefore, there are 2 positive and 2 negative roots of the function.<br />
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Example 3d.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 3rd degree, there are 3 zeros of the function.<br />There are 2 variation of signs in f(x) so there are either 2 positive roots or none at all.<br />Now, . There is only 1 variation of sign in f(x) so there is exactly 1 negative root.<br />Therefore, there are 2 positive roots and 1 negative root.<br />
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Example 3e.<br />Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />Solution:<br />Apply the Descartes’s Rule of Signs<br />Since f(x) is a polynomial of 5th degree, there are 5 zeros of the function.<br />There are 5 variations of signs in f(x) so there are either 5 or 3 positive roots.<br />Now, <br />There is no variation of sign in f(x) so there is no negative root.<br />Therefore, there are 5 positive roots of the function.<br />
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The Rational Zero Theorem<br />If the rational number p/q , a fraction in lowest terms, is a root of the equation <br />f(x) =an xn + an-1 xn-1+…+ a1 x + a0, <br /> where each a1 , a2 ,…, an is an integral coefficient, <br /> then p is an exact divisor of an and <br /> q is an exact divisor of a0 . <br />
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Example 4a.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Applying the Descartes’s Rule of Signs (see Example 3a), there are 3 negative roots. <br />By the Rational Zero Theorem, <br />p must be an exact divisor of 1 ; p = 1, -1<br />q must be an exact divisor of 1; q = 1, -1<br />the set of possible rational zeros p/q of f(x) is {1, -1}.<br />However, since there are 3 negative roots, we will only try x = -1. Perform synthetic division.<br />
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-1<br />-1<br />1<br />1<br />1 3 3 1 <br />-2<br />-1<br />-1<br />1<br />0<br />2<br />1 2 1 <br />-1<br />-1<br />1<br />0<br />Therefore, the zero of f(x) is 1 of multiplicity 3.<br />
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Example 4b.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Applying the Descartes’s Rule of Signs (see Example 3b), there are 1 positive and 2 negative roots. <br />By the Rational Zero Theorem, <br />p must be an exact divisor of -1 ; p = 1, -1<br />q must be an exact divisor of 1; q = 1, -1<br />the set of possible rational zeros p/q of f(x) is {1, -1}.<br />
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Example 4b.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Now use the values in the set of possible rational zeros to find the roots of the function.<br />Suppose x = 1. <br />Perform synthetic division.<br />
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1<br />-1<br />1<br />1<br />1 1 -1 -1 <br />2<br />1<br />1<br />1<br />0<br />2<br />1 2 1 <br />-1<br />-1<br />1<br />0<br />Therefore, the zeros of f(x) are -1, of multiplicity 2, and 1.<br />
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Example 4c.<br />Determine the zeros of the given polynomial function:<br />Solution:<br />Applying the Descartes’s Rule of Signs, there are 2 positive and 2 negative roots. <br />By the Rational Zero Theorem, <br />p must be an exact divisor of 4 ; p = 1, -1, 2, -2, 4, -4<br />q must be an exact divisor of 1; q = 1, -1<br />the set of possible rational zeros p/q of f(x) is<br /> {1, 2, 4}.<br />
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1<br />-1<br />1<br />1<br />1 0 -5 0 4 <br />-4<br />1<br />-4<br />1<br />0<br />1<br />-4<br />-4<br />1 1 -4 -4 <br />4<br />-1<br />0<br />-4<br />0<br />0<br />Note that <br /> So, <br />Therefore, the zeros of f(x) are 2, -2, 1, and -1. <br />
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Exercises:<br />1.Find the zeros of the given functions and determine the multiplicity of each zero.<br />1.1<br />1.2<br />1.3<br />2. Find a polynomial equation of least possible degree which satisfies the given conditions: <br />2.1 zeros are 1, -1, and ¾<br />2.2 Zeros are 4 of multiplicity 3, -1 of multiplicity 2, and 3<br />
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Exercises:<br />3. Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:<br />3.1<br />3.2<br />3.3<br />4. Determine the zeros of the given polynomial functions:<br />4.1 <br />4.2<br />4.3<br />
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