Q. 81 - 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.

$f\left(x\right)=x^2-5, [a,b]=[1,3]$

Step-by-Step Solution

Answer

For the function, $f\left(1\right)<0 \& f\left(3\right)>0$and the function is continuous, so the function must have at least one root between$x=1 \& x=3.$

The approximate root of the function is $x=\frac{47}{21}$

Step 1. Given information

The given function is $f\left(x\right)=x^2-5.$

Given interval is$[a,b]=[1,3].$

Step 2. Use Intermediate Value Theorem.

Determine function value at $x=1.$

$$\begin{gathered} f\left(x\right)=x^2-5 \\ f\left(1\right)=1^2-5 \\ f\left(1\right)=-4 \end{gathered}$$

Determine the function valuer at $x=3.$

$$\begin{gathered} f\left(x\right)=x^2-5 \\ f\left(3\right)=3^2-5 \\ f\left(3\right)=4 \end{gathered}$$

Here $f\left(1\right)<0 \& f\left(3\right)>0$and function is continuous.

so the function must have at least one root in the interval $[1,3].$

Step 3. approximation of root of function.

Differentiate the function.

$$\begin{gathered} \underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}=\underset{z\to x}{\lim }\frac{\left(z^2-5\right)-\left(x^2-5\right)}{z-x} \\ =\underset{z\to x}{\lim }\frac{z^2-x^2}{z-x} \\ =\underset{z\to x}{\lim }z+x \\ =2x \end{gathered}$$

Take $x_1=1 \& f\left(1\right)=4$for the approximation of the root.

Find derivative of the function at $x=1$

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

Determine the equation of a tangent to $f\left(x\right)$by using the points$\left(1,f\left(1\right)\right)=(1,-4).$

$$\begin{gathered} y-(-4)=2(x-1) \\ y=2x-6 \end{gathered}$$

Determine the roots of the tangent to $f\left(x\right).$

$$\begin{gathered} 0=2x-6 \\ x=3 \end{gathered}$$

Find the value of f at $x=3.$

$$\begin{gathered} f\left(3\right)=3^2-5 \\ f\left(3\right)=4 \end{gathered}$$

So $x=3$is not a root of function.

Step 4. approximation of root of function.

Take $x_2=3 \& f\left(3\right)=4$for the next approximation of the root.

Find derivative of the function at $x=3$

$$\begin{gathered} f\text{'}\left(3\right)=2\left(3\right) \\ f\text{'}\left(3\right)=6 \end{gathered}$$

Determine the equation of a tangent to function by using the points $(3,f(3\left)\right)=(3,4)$

$$\begin{gathered} y-4=6(x-3) \\ y=6x-14 \end{gathered}$$

Determine the roots of the tangent to the function

$$\begin{gathered} 0=6x-14 \\ x=\frac{14}{6}=\frac{7}{3} \end{gathered}$$

Find the value of the function at $x=\frac{7}{3}$

$$\begin{gathered} f\left(\frac{7}{3}\right)={\left(\frac{7}{3}\right)}^2-5 \\ f\left(\frac{7}{3}\right)=\frac{4}{9} \end{gathered}$$

So $x=\frac{7}{3}$is not a root of function.

Step 5. approximation of root of function.

Take $x=\frac{7}{3} \& f\left(\frac{7}{3}\right)=\frac{4}{9}$to approximate the root.

Find derivative of the function at $x=\frac{7}{3}.$

$$\begin{gathered} f\text{'}\left(\frac{7}{3}\right)=2\left(\frac{7}{3}\right) \\ f\text{'}\left(\frac{7}{3}\right)=\frac{14}{3} \end{gathered}$$

Determine the equation of a tangent to function by using the points $\left(\frac{7}{3},f\left(\frac{7}{3}\right)\right)=\left(\frac{7}{3},\frac{4}{9}\right).$

$$\begin{gathered} y-\frac{4}{9}=\frac{14}{3}\left(x-\frac{7}{3}\right) \\ y=\frac{14}{3}x-\frac{94}{9} \end{gathered}$$

Determine the roots of the tangent to the function

$$\begin{gathered} 0=\frac{14}{3}x-\frac{94}{9} \\ x=\frac{94}{42}=\frac{47}{21} \end{gathered}$$

Find the value of the function at $x=\frac{47}{21}.$

$$\begin{gathered} f\left(\frac{47}{21}\right)={\left(\frac{47}{21}\right)}^2-5 \\ f\left(\frac{47}{21}\right)=0.009 \end{gathered}$$

So is $x=\frac{47}{21}$not a root of function but the function value is very close to zero for this.

so approximate root is $x=\frac{47}{21}.$