91Ó°ÊÓ

We have given the following function :-

$f\left(x\right)=x^4-2,\left[a,b\right]=\left[1,2\right]$

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

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

The given function and interval is :-

$f\left(x\right)=x^4-2,\left[a,b\right]=\left[1,2\right]$

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 $f\left(x\right)=x^4-2$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 $\left[1,2\right]$.

For that firstly put $x=1$in the given function,. then we have :-

localid="1648369724468" $$\begin{gathered} f\left(1\right)=1^4-2 \\ ⇒f\left(1\right)=1-2 \\ ⇒f\left(1\right)=-1<0 \end{gathered}$$

Now put $x=2$in the given function, then we have :-

$$\begin{gathered} f\left(2\right)=2^4-2 \\ ⇒f\left(2\right)=16-2 \\ ⇒f\left(2\right)=14>0 \end{gathered}$$

We can see that $f\left(1\right)$and $f\left(2\right)$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 $\left[1,2\right]$.

The given function is :-

$f\left(x\right)=x^4-2$

Then derivative of this function is as following :-

$f\text{'}\left(x\right)=4x^3$.

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

Let the initial approximation is :-

role="math" localid="1648370775632" $x_0=1$

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 :-

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

Put all the values, then we have :-

$$\begin{gathered} x_1=1-\frac{1^4-2}{4\left(1^3\right)} \\ ⇒x_1=1-\frac{-1}{4} \\ ⇒x_1=1-(-0.25) \\ ⇒x_1=1.25 \end{gathered}$$

So the first approximation to the root is$1.25$

The first approximation to the root is $x_1=1.25$. 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 :-

$$\begin{gathered} x_2=1.25-\frac{{(1.25)}^4-2}{4{(1.25)}^3} \\ ⇒x_2=1.25-\frac{0.4414}{7.8125} \\ ⇒x_2=1.25-0.0565 \\ ⇒x_2=1.1935 \end{gathered}$$

So the second approximation to the root is$1.1935$.

The second approximation to the root is $x_2=1.1935$. Then the third approximation to the root is :-

$x_3=x_2-\frac{f\left(x_2\right)}{f\text{'}\left(x_2\right)}$

Put all the values, then we have :-

$$\begin{gathered} x_3=1.1935-\frac{{(1.1935)}^4-2}{4{(1.1935)}^3} \\ ⇒x_3=1.1935-\frac{0.029036}{6.8003} \\ ⇒x_3=1.1935-0.00427 \\ ⇒x_3=1.18923 \end{gathered}$$

So the third approximation to the root is$1.18923$.

The third approximation to the root is $x_3=1.18923$. 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 :-

$$\begin{gathered} x_4=1.18923-\frac{{(1.18923)}^4-2}{4{(1.18923)}^3} \\ ⇒x_4=1.18923-\frac{0.000154}{6.72756} \\ ⇒x_4=1.18923-0.0000229 \\ ⇒x_4=1.189207 \end{gathered}$$

We can see that the third and fourth approximations are same up to four decimal places.

So that the required approximation to the root is$1.1892$.