SOLUTION: 3) Study the method of linear interpolation presented in chapter 5. Given p(x) = 2x^3 – 3x^2 – 6x + 4, show that there is an irrational root in the interval (0,1) and use linear in

Algebra ->  Rational-functions -> SOLUTION: 3) Study the method of linear interpolation presented in chapter 5. Given p(x) = 2x^3 – 3x^2 – 6x + 4, show that there is an irrational root in the interval (0,1) and use linear in      Log On


   

Question 133510: 3) Study the method of linear interpolation presented in chapter 5. Given p(x) = 2x^3 – 3x^2 – 6x + 4, show that there is an irrational root in the interval (0,1) and use linear interpolation to find the root accurate to two decimal places.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First let's find the possible rational roots of f%28x%29=2x%5E3-3x%5E2-6x+%2B+4


Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients


So let's list the factors of 4 (the last coefficient):



Now let's list the factors of 2 (the first coefficient):



Now let's divide each factor of the last coefficient by each factor of the first coefficient









Now simplify

These are all the distinct rational zeros of the function that could occur




So the only possible rational roots in the interval (0,1) is 1%2F2. However, if we plug in f%281%2F2%29, we get


f%281%2F2%29=2%281%2F2%29%5E3-3%281%2F2%29%5E2-6%281%2F2%29+%2B+4


f%281%2F2%29=1%2F2


So we can see that 1%2F2 is not a zero of f%28x%29=2x%5E3-3x%5E2-6x+%2B+4. So there are no rational roots in the interval (0,1)



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Now let's see if there is a zero in the interval (0,1)



Let's evaluate f(0) (the left endpoint of the interval)


f%28x%29=2x%5E3-3x%5E2-6x%2B4 Start with the given function


f%280%29=2%280%29%5E3-3%280%29%5E2-6%280%29%2B4 Plug in x=0


f%280%29=2%2A0-3%2A0%5E2-6%2A0%2B4 Raise 0 to the 3rd power to get 0


f%280%29=0-3%2A0%5E2-6%2A0%2B4 Multiply 2 and 0 to get 0


f%280%29=0-3%2A0-6%2A0%2B4 Raise 0 to the 2nd power to get 0


f%280%29=0-0-6%2A0%2B4 Multiply 3 and 0 to get 0


f%280%29=0-6%2A0%2B4 Subtract 0 from 0 to get 0


f%280%29=0-0%2B4 Multiply 6 and 0 to get 0


f%280%29=0%2B4 Subtract 0 from 0 to get 0


f%280%29=4 Add 0 and 4 to get 4


So when x=0, we have y=4 (notice how y is positive)



Now let's evaluate f(1) (the right endpoint of the interval)




f%28x%29=2x%5E3-3x%5E2-6x%2B4 Start with the given function


f%281%29=2%281%29%5E3-3%281%29%5E2-6%281%29%2B4 Plug in x=1


f%281%29=2%2A1-3%2A1%5E2-6%2A1%2B4 Raise 1 to the 3rd power to get 1


f%281%29=2-3%2A1%5E2-6%2A1%2B4 Multiply 2 and 1 to get 2


f%281%29=2-3%2A1-6%2A1%2B4 Raise 1 to the 2nd power to get 1


f%281%29=2-3-6%2A1%2B4 Multiply 3 and 1 to get 3


f%281%29=-1-6%2A1%2B4 Subtract 3 from 2 to get -1


f%281%29=-1-6%2B4 Multiply 6 and 1 to get 6


f%281%29=-7%2B4 Subtract 6 from -1 to get -7


f%281%29=-3 Add -7 and 4 to get -3


So when x=1, we have y=-3 (notice how y is negative)



Since the sign of y changes from positive to negative as x goes from 0 to 1, this means that there must be a zero in the interval (0,1)


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Conclusion:


Since we know that there is a zero in the interval (0,1), but it is not a rational zero, this means that the zero must be irrational.


So there is an irrational zero in the interval (0,1)




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Now let's use linear interpolation to find the equation of the line that goes through 2 points on the curve


Since we previously found that f%280%29=4 and f%281%29=-3, we have the two points (0,4) and (1,-3) that lie on the line.


Now let's find the equation of the line through (0,4) and (1,-3)


First lets find the slope through the points (0,4) and (1,-3)

m=%28y%5B2%5D-y%5B1%5D%29%2F%28x%5B2%5D-x%5B1%5D%29 Start with the slope formula (note: is the first point (0,4) and is the second point (1,-3))

m=%28-3-4%29%2F%281-0%29 Plug in y%5B2%5D=-3,y%5B1%5D=4,x%5B2%5D=1,x%5B1%5D=0 (these are the coordinates of given points)

m=+-7%2F1 Subtract the terms in the numerator -3-4 to get -7. Subtract the terms in the denominator 1-0 to get 1


m=-7 Reduce

So the slope is
m=-7

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Now let's use the point-slope formula to find the equation of the line:



------Point-Slope Formula------
y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope, and is one of the given points

So lets use the Point-Slope Formula to find the equation of the line

y-4=%28-7%29%28x-0%29 Plug in m=-7, x%5B1%5D=0, and y%5B1%5D=4 (these values are given)


y-4=-7x%2B%28-7%29%280%29 Distribute -7

y-4=-7x%2B0 Multiply -7 and 0 to get 0%2F0. Now reduce 0%2F0 to get 0

y=-7x%2B0%2B4 Add 4 to both sides to isolate y

y=-7x%2B4 Combine like terms 0 and 4 to get 4


So we have the line y=-7x%2B4 that goes through (0,4) and (1,-3) which lie on the curve.



0=-7x%2B4 Now plug in y=0




-4=-7x Subtract 4 from both sides


-4%2F%28-7%29=x Divide both sides by 7 to isolate x







So our answer is approximately x=0.571428571428571


So the root to two decimal places is x=0.57