Q. 83 - 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.

$f\left(x\right)=x^3-3x+1, [a,b]=[0,1]$

Step-by-Step Solution

Answer

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

The approximate root of the function is $x=\frac{25}{72}.$

Step 1. Given information

The given function is $f\left(x\right)=x^3-3x+1.$

Given interval is $[a,b]=[0,1].$

Step 2. Use Intermediate Value Theorem.

Determine function value at $x=0$

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

Determine the function valuer at $x=1$

$$\begin{gathered} f\left(1\right)=1^3-3\left(1\right)+1 \\ f\left(1\right)=-1 \end{gathered}$$

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

so the function must have at least one root in the given interval.

Step 3. approximation of root of function.

Differentiate the function.

$$\begin{gathered} \underset{z\to x}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x}=\underset{z\to x}{\lim }\frac{\left(z^3-3z+1\right)-\left(x^3-3x+1\right)}{z-x} \\ =\underset{z\to x}{\lim }\frac{(z-x)\left(z^2+zx+x^2\right)-3(z-x)}{z-x} \\ =\underset{z\to x}{\lim }{z}^2+zx+x^2-3 \\ =3x^2-3 \end{gathered}$$

Take $x=0 \& f\left(0\right)=1$for the approximation of the root.

Find derivative of the function at $x=0.$

$$\begin{gathered} f\text{'}\left(0\right)=3{\left(0\right)}^2-3 \\ f\text{'}\left(0\right)=-3 \end{gathered}$$

Determine the equation of a tangent to function by using the points $\left(0,f\left(0\right)\right)=(0,1).$

$$\begin{gathered} y-1=-3(x-0) \\ y=-3x+1 \end{gathered}$$

Determine the roots of the tangent to function.

$$\begin{gathered} 0=-3x+1 \\ x=\frac{1}{3} \end{gathered}$$

Find the value of the function at $x=\frac{1}{3}.$

$$\begin{gathered} f\left(\frac{1}{3}\right)={\left(\frac{1}{3}\right)}^3-3\left(\frac{1}{3}\right)+1 \\ f\left(\frac{1}{3}\right)=\frac{1}{27} \end{gathered}$$

So $x=\frac{1}{3}$is not a root of function.

Step 4. approximation of root of function.

Take $x=\frac{1}{3} \& f\left(\frac{1}{3}\right)=\frac{1}{27}$for the approximation of the root.

Find derivative of the function at $x=\frac{1}{3}$

$$\begin{gathered} f\text{'}\left(\frac{1}{3}\right)=3{\left(\frac{1}{3}\right)}^2-3 \\ f\text{'}\left(\frac{1}{3}\right)=\frac{-8}{3} \end{gathered}$$

Determine the equation of a tangent to function by using the points$\left(\frac{1}{3},f\left(\frac{1}{3}\right)\right)=\left(\frac{1}{3},\frac{1}{27}\right)$

$$\begin{gathered} y-\frac{1}{27}=\frac{-8}{3}\left(x-\frac{1}{3}\right) \\ y=\frac{-8}{3}x+\frac{25}{27} \end{gathered}$$

Determine the roots of the tangent to function.

$$\begin{gathered} 0=\frac{-8}{3}x+\frac{25}{27} \\ x=\frac{25}{27}\times \frac{3}{8} \\ x=\frac{25}{72}. \end{gathered}$$

Find the value of the function at $x=\frac{25}{72}$

$$\begin{gathered} f\left(\frac{25}{72}\right)={\left(\frac{25}{72}\right)}^3-3\left(\frac{25}{72}\right)+1 \\ f\left(\frac{25}{72}\right)=\frac{73}{373242} \\ f\left(\frac{25}{72}\right)=0.00019. \end{gathered}$$

 $x=\frac{25}{72}$is not a root of function but it is close to the real root.

so the approximate root of the function is $x=\frac{25}{72}.$