Answer to Question #313160 in Calculus for Izy

Question

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

Answer 1

Solution

The given function is

$f(x) = x³ - 3x + 5$

$f(-4) = (-4)³ - 3(-4) + 5 = -47<0$

$f(0) = (0)³ - 3(0) + 5 = 5>0$

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

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

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

$f(-3) = (-3)³ - 3(-3) + 5 = -13<0$

$f(-1) = (-1)³ - 3(-1) + 5 =7 >0$

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

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

Now consider,

$f(-2) = (-2)³ - 3(-2) + 5 =3 >0$

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

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

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 $f(x) = x³ - 3x + 5$ , which entirely ensures that the root lies within the interval [−3,−2]

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