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

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) (Show Source): You can put this solution on YOUR website!
First let's find the possible rational roots of

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 . However, if we plug in , we get

So we can see that is not a zero of . 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)

Start with the given function

Plug in

Raise 0 to the 3rd power to get 0

Multiply 2 and 0 to get 0

Raise 0 to the 2nd power to get 0

Multiply 3 and 0 to get 0

Subtract 0 from 0 to get 0

Multiply 6 and 0 to get 0

Subtract 0 from 0 to get 0

Add 0 and 4 to get 4

So when , we have (notice how y is positive)

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

Start with the given function

Plug in

Raise 1 to the 3rd power to get 1

Multiply 2 and 1 to get 2

Raise 1 to the 2nd power to get 1

Multiply 3 and 1 to get 3

Subtract 3 from 2 to get -1

Multiply 6 and 1 to get 6

Subtract 6 from -1 to get -7

Add -7 and 4 to get -3

So when , we have (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)

--------------------------

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 and , 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 (,) and (,)

Start with the slope formula (note:

is the first point (,) and

is the second point (,))

Plug in ,,, (these are the coordinates of given points)

Subtract the terms in the numerator to get . Subtract the terms in the denominator to get

Reduce

So the slope is

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

------Point-Slope Formula------
where is the slope, and

is one of the given points

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

Plug in , , and (these values are given)

Distribute

Multiply and to get . Now reduce to get

Add to both sides to isolate y

Combine like terms and to get

So we have the line that goes through (0,4) and (1,-3) which lie on the curve.

Now plug in

Subtract 4 from both sides

Divide both sides by 7 to isolate x

So our answer is approximately

So the root to two decimal places is

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