 Section 5.5 – The Real Zeros of a Rational Function

Presentation on theme: "Section 5.5 – The Real Zeros of a Rational Function"— Presentation transcript:

Section 5.5 – The Real Zeros of a Rational Function
Remainder Theorem If f(x) is a polynomial function and is divided by x – c, then the remainder is f(c). Example: 𝑓 𝑥 = 𝑥 2 −2𝑥−15 𝐷𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟: 𝑥−4 𝑜𝑟 𝑥=4 𝑓 4 = 4 2 −2 4 −15 𝑓 4 =−7 The remainder after dividing f(x) by (x – 4) would be -7. 4 8 1 2 −7

Section 5.5 – The Real Zeros of a Rational Function
Factor Theorem If f(x) is a polynomial function, then x – c is a factor of f(x) if and only if f(c) = 0. Example: 𝑓 𝑥 = 𝑥 2 −2𝑥−15 𝐷𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑡ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟: 𝑥+3 𝑜𝑟 𝑥=−3 𝑓 −3 = (−3) 2 −2 −3 −15 𝑓 −3 =0 The remainder after dividing f(x) by (x + 3) would be 0. −3 15 1 −5

Section 5.5 – The Real Zeros of a Rational Function
Rational Zeros Theorem (for functions of degree 1 or higher) Given: (1) 𝑓 𝑥 = 𝑎 𝑛 𝑥 𝑛 + 𝑎 𝑛−1 𝑥 𝑛−1 + ⋯ 𝑎 1 𝑥+ 𝑎 0 (2) Each coefficient is an integer. If 𝑝 𝑞 (in lowest terms) is a rational zero of the function, then p is a factor of 𝑎 0 and q is a factor of 𝑎 𝑛 . Theorem: A polynomial function of odd degree with real coefficients has at least one real zero.

Section 5.5 – The Real Zeros of a Rational Function
Rational Zeros Theorem Example: Find the solution(s) of the equation. 𝑓 𝑥 = 𝑥 3 −2 𝑥 2 −5𝑥+6 𝑝: ±1, ±2, ±3, ± 𝑞: ±1 𝑝 𝑞 : ± 1 1 , ± 2 1 , ± 3 1 , ± 6 1 Possible solutions: 𝑥=±1, ±2, ±3, ±6 Try: 𝑥= 𝑜𝑟 𝑥−1=0

Section 5.5 – The Real Zeros of a Rational Function
𝑓 𝑥 = 𝑥 3 −2 𝑥 2 −5𝑥+6 Long Division Synthetic Division 𝑥 2 −𝑥 −6 𝑥 3 −𝑥 2 1 −1 −6 −𝑥 2 −5𝑥 1 −1 −6 −𝑥 2 +𝑥 −6𝑥 +6 𝑥−1 𝑥 2 −𝑥−6 −6𝑥 +6 𝑥−1 𝑥 2 −𝑥−6

Section 5.5 – The Real Zeros of a Rational Function
𝑓 𝑥 = 𝑥 3 −2 𝑥 2 −5𝑥+6 𝑥−1 𝑥 2 −𝑥−6 =0 𝑥−1 𝑥+2 𝑥−3 =0 𝑥−1 =0 𝑥+2 =0 𝑥−3 =0 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠: 𝑥=−2, 1, 3

Section 5.5 – The Real Zeros of a Rational Function
Example: Find the solution(s) of the equation. 𝑓 𝑥 = 4𝑥 4 +7 𝑥 2 −2 𝑝: ±1, ± 𝑞: ±1, ±2, ±4 Possible solutions ( 𝑝 𝑞 ): 𝑥=±1, ± 1 2 , ± 1 4 , ±2 Try: 𝑥=1 Try: 𝑥=2 4 4 11 11 8 16 46 92 4 4 11 11 9 4 8 23 46 90 Try: 𝑥= 1 2 (𝑥− 1 2 )(4 𝑥 3 +2 𝑥 2 +8𝑥+4) 2 1 4 2 4 2 8 4

Section 5.5 – The Real Zeros of a Rational Function
(𝑥− 1 2 )(4 𝑥 3 +2 𝑥 2 +8𝑥+4) (𝑥− 1 2 )(2)(2 𝑥 3 + 𝑥 2 +4𝑥+2) (𝑥− 1 2 )(2)( 𝑥 2 2𝑥 𝑥+1 ) (𝑥− 1 2 )(2)( 𝑥 2 +2) 2𝑥+1 𝑥− 1 2 = 𝑥 2 +2= 𝑥+1=0 𝑥=− 1 2 , 1 2 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠:

Section 5.5 – The Real Zeros of a Rational Function
Intermediate Value Theorem In a polynomial function, if a < b and f(a) and f(b) are of opposite signs, then there is at least one real zero between a and b. (𝑏, 𝑓 𝑏 ) (𝑎, 𝑓 𝑎 ) 𝑟𝑒𝑎𝑙 𝑧𝑒𝑟𝑜 𝑟𝑒𝑎𝑙 𝑧𝑒𝑟𝑜 (𝑏, 𝑓 𝑏 ) (𝑎, 𝑓 𝑎 )

Section 5.5 – The Real Zeros of a Rational Function
Intermediate Value Theorem Do the following polynomial functions have at least one real zero in the given interval? 𝑓 𝑥 =2 𝑥 3 −3 𝑥 2 −2 𝑓 𝑥 =2 𝑥 3 −3 𝑥 2 −2 [0, 2] [3, 6] 𝑓 0 = −2 𝑓 2 = 2 𝑓 3 = 25 𝑓 6 = 322 𝑦𝑒𝑠 𝑛𝑜𝑡 𝑒𝑛𝑜𝑢𝑔ℎ 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑓 𝑥 = 𝑥 4 −2 𝑥 2 −3𝑥−3 𝑓 𝑥 = 𝑥 4 −2 𝑥 2 −3𝑥−3 [−5, −2] [−1, 3] 𝑓 −5 = 587 𝑓 −2 = 11 𝑓 −1 = −1 𝑓 3 = 51 𝑛𝑜𝑡 𝑒𝑛𝑜𝑢𝑔ℎ 𝑖𝑛𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑦𝑒𝑠

Download ppt "Section 5.5 – The Real Zeros of a Rational Function"

Similar presentations