Q. 82 - Calculus

Question

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

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

Step-by-Step Solution

Answer

For the function, $f\left(1\right)<0 \& f\left(2\right)>0$and the function is continuous, so the function must have at least one root between$x=1 \& x=2.$

The approximate root of the function is $x=1.417.$

Step 1. Given information

The given function is $f\left(x\right)=x^2-2.$

Given interval is $[a,b]=[1,2].$

Step 2. Use Intermediate Value Theorem.

Determine function value at $x=1.$

$$\begin{gathered} f\left(1\right)=1^2-2 \\ f\left(1\right)=-1 \end{gathered}$$

Determine the function valuer at $x=2.$

$$\begin{gathered} f\left(2\right)=2^2-2 \\ f\left(2\right)=2 \end{gathered}$$

Here $f\left(1\right)<0 \& f\left(2\right)>0$and function is continuous.

so the function must have at least one root in the interval $[1,2]$

Step 3. approximation of root of function.

Differentiate the function.

$$\begin{gathered} \underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}=\underset{z\to x}{\lim }\frac{\left(z^2-2\right)-\left(x^2-2\right)}{z-x} \\ =\underset{z\to x}{\lim }\frac{z^2-x^2}{z-x} \\ =\underset{z\to x}{\lim }z+x \\ =2x \end{gathered}$$

Take $x=1 \& f\left(1\right)=-1$to approximate the root.

Find derivative of the function at $x=1.$

$$\begin{gathered} f\text{'}\left(1\right)=2\left(1\right) \\ f\text{'}\left(1\right)=2 \end{gathered}$$

Determine the equation of a tangent to function by using the points$\left(1,f\left(1\right)\right)=(1,-1).$

$$\begin{gathered} y-(-1)=2(x-1) \\ y=2x-3 \end{gathered}$$

Determine the roots of the tangent to function.

$$\begin{gathered} 0=2x-3 \\ x=\frac{3}{2}=1.5 \end{gathered}$$

Find the value of the function at $x=1.5$

$$\begin{gathered} f(1.5)=1.5^2-5 \\ f(1.5)=0.25 \end{gathered}$$

So $x=1.5$is not a root of function.

Step 4. approximation of root of function.

Take $x=1.5 \& f(1.5)=0.25$for the approximation of the root.

Find derivative of the function at $x=1.5$

$$\begin{gathered} f\text{'}(1.5)=2(1.5) \\ f\text{'}(1.5)=3 \end{gathered}$$

Determine the equation of a tangent to function by using the points$(1.5,f(1.5\left)\right)=(1.5,0.25)$

$$\begin{gathered} y-0.25=3(x-1.5) \\ y=3x-4.25 \end{gathered}$$

Determine the roots of the tangent to function.

$$\begin{gathered} 0=3x-4.25 \\ x=1.417 \end{gathered}$$

Find the value of the function at  $x=1.417$

$$\begin{gathered} f(1.417)={\left(1.42\right)}^2-2 \\ f(1.417)=0.007 \end{gathered}$$

So $x=1.417$is not a root of function but it is very close to the real root.

so approximate root is $x=1.417.$