Question 133510
First let's find the possible rational roots of {{{f(x)=2x^3-3x^2-6x + 4}}}



Any rational zero can be found through this equation


*[Tex \LARGE Roots=\frac{p}{q}] where p and q are the factors of the last and first coefficients



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


*[Tex \LARGE p=\pm1, \pm2, \pm4]


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


*[Tex \LARGE q=\pm1, \pm2]


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



*[Tex \LARGE \frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{2}{2}, \frac{4}{1}, \frac{4}{2}, \frac{-1}{1}, \frac{-1}{2}, \frac{-2}{1}, \frac{-2}{2}, \frac{-4}{1}, \frac{-4}{2}]







Now simplify


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


*[Tex \LARGE  1, \frac{1}{2}, 2, 4, -1, \frac{-1}{2}, -2, -4]



So the only possible rational roots in the interval (0,1) is {{{1/2}}}. However, if we plug in {{{f(1/2)}}}, we get 



{{{f(1/2)=2(1/2)^3-3(1/2)^2-6(1/2) + 4}}}



{{{f(1/2)=1/2}}}



So we can see that {{{1/2}}} is <b>not</b> a zero of {{{f(x)=2x^3-3x^2-6x + 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(x)=2x^3-3x^2-6x+4}}} Start with the given function



{{{f(0)=2(0)^3-3(0)^2-6(0)+4}}} Plug in {{{x=0}}}



{{{f(0)=2*0-3*0^2-6*0+4}}} Raise 0 to the 3rd power to get 0



{{{f(0)=0-3*0^2-6*0+4}}} Multiply 2 and 0 to get 0



{{{f(0)=0-3*0-6*0+4}}} Raise 0 to the 2nd power to get 0



{{{f(0)=0-0-6*0+4}}} Multiply 3 and 0 to get 0



{{{f(0)=0-6*0+4}}} Subtract 0 from 0 to get 0



{{{f(0)=0-0+4}}} Multiply 6 and 0 to get 0



{{{f(0)=0+4}}} Subtract 0 from 0 to get 0



{{{f(0)=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(x)=2x^3-3x^2-6x+4}}} Start with the given function



{{{f(1)=2(1)^3-3(1)^2-6(1)+4}}} Plug in {{{x=1}}}



{{{f(1)=2*1-3*1^2-6*1+4}}} Raise 1 to the 3rd power to get 1



{{{f(1)=2-3*1^2-6*1+4}}} Multiply 2 and 1 to get 2



{{{f(1)=2-3*1-6*1+4}}} Raise 1 to the 2nd power to get 1



{{{f(1)=2-3-6*1+4}}} Multiply 3 and 1 to get 3



{{{f(1)=-1-6*1+4}}} Subtract 3 from 2 to get -1



{{{f(1)=-1-6+4}}} Multiply 6 and 1 to get 6



{{{f(1)=-7+4}}} Subtract 6 from -1 to get -7



{{{f(1)=-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 <b>not</b> 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(0)=4}}} and {{{f(1)=-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=(y[2]-y[1])/(x[2]-x[1])}}} Start with the slope formula (note: *[Tex \Large \left(x_{1},y_{1}\right)] is the first point ({{{0}}},{{{4}}}) and  *[Tex \Large \left(x_{2},y_{2}\right)] is the second point ({{{1}}},{{{-3}}}))


{{{m=(-3-4)/(1-0)}}} Plug in {{{y[2]=-3}}},{{{y[1]=4}}},{{{x[2]=1}}},{{{x[1]=0}}}  (these are the coordinates of given points)


{{{m= -7/1}}} 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[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(\textrm{x_{1},y_{1}}\right)] is one of the given points


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


{{{y-4=(-7)(x-0)}}} Plug in {{{m=-7}}}, {{{x[1]=0}}}, and {{{y[1]=4}}} (these values are given)



{{{y-4=-7x+(-7)(0)}}} Distribute {{{-7}}}


{{{y-4=-7x+0}}} Multiply {{{-7}}} and {{{0}}} to get {{{0/0}}}. Now reduce {{{0/0}}} to get {{{0}}}


{{{y=-7x+0+4}}} Add {{{4}}} to  both sides to isolate y


{{{y=-7x+4}}} Combine like terms {{{0}}} and {{{4}}} to get {{{4}}} 



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




{{{0=-7x+4}}} Now plug in {{{y=0}}}





{{{-4=-7x}}} Subtract 4 from both sides



{{{-4/(-7)=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}}}