Question #313160

Find the distinct interval of length 1 containing a root or solutin of f(x) = x³ - 3x + 5 using IVT

1
Expert's answer
2022-03-18T12:33:05-0400

Solution


The given function is


f(x)=x33x+5f(x) = x³ - 3x + 5


f(4)=(4)33(4)+5=47<0f(-4) = (-4)³ - 3(-4) + 5 = -47<0


f(0)=(0)33(0)+5=5>0f(0) = (0)³ - 3(0) + 5 = 5>0


Since f(0)>0f(0)>0 and f(4)<0f(-4)<0 , therefore, the root of the given function lies between the interval [4,0][-4, 0]


The length of this interval [4,0][-4, 0] is =(0)(4)=4=(0)-(-4)=4 .


We narrow this interval, such that we get the length of the interval equal to 1.


f(3)=(3)33(3)+5=13<0f(-3) = (-3)³ - 3(-3) + 5 = -13<0


f(1)=(1)33(1)+5=7>0f(-1) = (-1)³ - 3(-1) + 5 =7 >0


Again we see that f(1)>0f(-1)>0 and f(3)<0f(-3)<0 , therefore, the root of the given function lies between the interval [3,1][-3, -1]


The length of this interval [3,1][-3, -1] is =(1)(3)=2=(-1)-(-3)=2 .


Now consider,


f(2)=(2)33(2)+5=3>0f(-2) = (-2)³ - 3(-2) + 5 =3 >0


Still we can see that t f(2)>0f(-2)>0 and f(3)<0f(-3)<0 , therefore, the root of the given function lies between the interval [3,2][-3, -2]


And the length of the interval is =(2)(3)=1=(-2)-(-3)=1.



Hence the required interval of length 1, where the root of the given function lies is [−3,−2]



The diagram below is the plot of the given function f(x)=x33x+5f(x) = x³ - 3x + 5 , which entirely ensures that the root lies within the interval [−3,−2]





The root is 2.279-2.279, which is within the interval [3,2][-3,-2]



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