Q 85 - Calculus

Question

For each function $f\left(x\right)$ and interval $\left[a,b\right]$ in Exercises $81-86$, use the Intermediate Value Theorem to argue that the function must have at least one real root on $\left[a,b\right]$. Then apply Newton’s method to approximate that root.

$f\left(x\right)=x^3+1,\left[a,b\right]=\left[-2,1\right]$

Step-by-Step Solution

Answer

By using intermediate value theorem, it is clear that there exists a root of the given function $f\left(x\right)=x^3+1$ on the interval  $\left[-2,1\right]$.

By using Newton's Method we find that the approximated root of the given function lies in that interval is $-1.000 or -1$.

Step 1. Given Information

We have given the following function :-

$f\left(x\right)=x^3+1,\left[a,b\right]=\left[-2,1\right]$

Firstly we have to use intermediate value theorem to show that one real root of this given function lies between the given interval $\left[-2,1\right]$.

After that we have to apply Newton's Method to find the approximate value of that root. 

Step 2. To show that one real root lies between the given Interval

The given function and interval is :- 

$f\left(x\right)=x^3+1,\left[a,b\right]=\left[-2,1\right]$.

We know that Intermediate value theorem, stated that If $f\left(x\right)$ is a continuous function has values of opposite sign inside an interval $\left[a,b\right]$, then it has a real root in that interval.

Here we have $f\left(x\right)=x^3+1$ is a polynomial function and every polynomial is a continuous function.

So the given function is a continuous function.

Now check the values of function at the end points of the given interval $\left[-2,1\right]$.

For that firstly put $x=-2$ in the given function,. then we have :-

$$\begin{gathered} f(-2)={\left(-2\right)}^3+1 \\ \Rightarrow f(-2)=-8+1 \\ \Rightarrow f(-2)=-7<0 \end{gathered}$$

Now put $x=1$ in the given function, then we have :-

$$\begin{gathered} f\left(1\right)=1^3+1 \\ \Rightarrow f\left(1\right)=1+1 \\ \Rightarrow f\left(1\right)=2>0 \end{gathered}$$

We can see that $f(-2)$ and $f\left(1\right)$ has values of opposite sign. Then we can conclude by using intermediate value theorem, that there exist at least one real root of the given function on the interval $\left[-2,1\right]$. 

Step 3. First approximation to root by applying Newton's Method

The given function is :- 

$f\left(x\right)=x^3+1$

Then derivative of this function is as following :- 

$f\text{'}\left(x\right)=3x^2$.

Now to apply newton's method, we have to let an initial approximation.

Let the initial approximation is :-

$x_0=-2$.

Now the formula of newton's method is :- 

$x_n=x_{n-1}-\frac{f\left(x_{n-1}\right)}{f\text{'}\left(x_{n-1}\right)}$

Then the first approximation to root is :- 

$x_1=x_0-\frac{f(-2)}{f\text{'}(-2)}$

Put all the values, then we have :- 

$$\begin{gathered} x_1=-2-\frac{{(-2)}^3+1}{3{(-2)}^2} \\ \Rightarrow x_1=-2-\frac{-8+1}{3\left(4\right)} \\ \Rightarrow x_1=-2-\frac{-7}{12} \\ \Rightarrow x_1=-2-(-0.58334) \\ \Rightarrow x_1=-1.41666 \end{gathered}$$

So the first approximation to the root is  $-1.41666$.

Step 4. Second approximation to the root by applying Newton's Method

The first approximation to the root is $x_1=-1.41666$. Then the second approximation to the root is :-

$x_2=x_1-\frac{f\left(x_1\right)}{f\text{'}\left(x_1\right)}$

Put all the values, then we have :- 

$$\begin{gathered} x_2=-1.41666-\frac{{(-1.41666)}^3+1}{3{(-1.41666)}^2} \\ \Rightarrow x_2=-1.41666-\frac{-2.84313+1}{3(2.007)} \\ \Rightarrow x_2=-1.41666-\frac{-1.84313}{6.021} \\ \Rightarrow x_2=-1.41666-(-0.306117) \\ \Rightarrow x_2=-1.110543 \end{gathered}$$

So the second approximation to the root is $-1.110543$.

Step 5. Third approximation to root by applying Newton's Method

The second approximation to the root is $x_3=-1.110543$. Then the third approximation to the root is :-

$$\begin{gathered} x_3=x_2-\frac{f\left(x_2\right)}{f\text{'}\left(x_2\right)} \\ \Rightarrow x_3=-1.110543-\frac{{(-1.110543)}^3+1}{3{(-1.110543)}^2} \\ \Rightarrow x_3=-1.110543-\frac{-1.36934+1}{3(1.233306)} \\ \Rightarrow x_3=-1.110543-\frac{-0.36934}{3.69992} \\ \Rightarrow x_3=-1.110543-(-0.099824) \\ \Rightarrow x_3=-1.010719 \end{gathered}$$

So the third approximation to the root is $-1.010719$.

Step 6. Fourth approximation to root by applying Newton's Method

The third approximation to the root is $x_3=-1.010719$. Then the fourth approximation to the root is :-

$x_4=x_3-\frac{f\left(x_3\right)}{f\text{'}\left(x_3\right)}$

Put all the values, then we have :- 

$$\begin{gathered} x_4=-1.010719-\frac{{(-1.010719)}^3+1}{3{(-1.010719)}^2} \\ \Rightarrow x_4=-1.010719-\frac{-1.0325+1}{3.19819} \\ \Rightarrow x_4=-1.010719-\frac{-0.0325}{3.19819} \\ \Rightarrow x_4=-1.010719-(-0.01) \\ \Rightarrow x_4=1.000719 \end{gathered}$$

Step 7. Fifth approximation to root by applying Newton's Method

The fourth approximation to the root is $x_4=-1.000719$. Then the fifth approximation to the root is :-

$x_5=x_4-\frac{f\left(x_4\right)}{f\text{'}\left(x_4\right)}$

Put all the values, then we have :- 

$$\begin{gathered} x_5=-1.000719-\frac{{(-1.000719)}^3+1}{3{(-1.000719)}^2} \\ \Rightarrow x_5=-1.000719-\frac{-1.00216+1}{3.004316} \\ \Rightarrow x_5=-1.000719-\frac{-0.00216}{3.004316} \\ \Rightarrow x_5=-1.000719-(-0.0007189) \\ \Rightarrow x_5=-1.0000001 \end{gathered}$$

We can see that the fourth and fifth approximations are same up to three decimal places.

So that the required approximation to the root is $-1.000$ or $-1$.