Chapter 5, Problem 4P

Determine the roots of $f\left(x\right)=-12-21x+18x^2-2.75x^3$ graphically. In addition, determine the first root of the function with (b) bisection, and (c) false position. For (b) and (c) use initial guesses of $x_l=-1\text{and }{x}_u=0$, and a stopping criterion of 1%.

To determine

To calculate: The real roots of the equation $f\left(x\right)=-2.75x^3+18x^2-21x-12$ using the graphical method.

Answer to Problem 4P

Solution: Graphically, the real roots of the equation are approximated as $-0.4,2.2,\text{ and 4}\text{.8}$.

Explanation of Solution

Given Information: The equation $f\left(x\right)=-25+82x-90x^2+44x^3-8x^4+0.7x^5$.

Formula used: The roots of an equation can be represented graphically by the x-coordinate of the point where the graph intercepts the x-axis.

Calculation:

Use the following MATLAB code to plot the function $f\left(x\right)$ with respect to independent variable $x$ as,

% (a)

clear;clc;

% define the vector x

x = linspace(-1,5);

% define the function f(x)

y1 = -2.75*x.^3+18*x.^2-21*x-12;

y2 = 0;

% plot f(x)

plot(x, y1);

xlabel('x')

ylabel('f(x)')

grid on

hold on

line([-1,5],[y2, y2])

Execute the above code to obtain the plot as,

From the plot, the zeros of the equation are approximated as $-0.4$, 2.2 and 4.8.

To determine

To calculate: The root of the equation $f\left(x\right)=-2.75x^3+18x^2-21x-12$ using the bisection method with $x_l$ and $x_u$ as $-1$ and 0, respectively, and with the stopping criteria being the iteration where ${\epsilon }_a$ falls below 1%.

Answer to Problem 4P

Solution: The root of the equation is approximated as $-0.41797$.

Explanation of Solution

Given Information: The function equation is $f\left(x\right)=-2.75x^3+18x^2-21x-12$.

Initial guesses are,

$\begin{array}{l}{x}_l=-1\\ x_u=0\end{array}$

Formula used: A root of an equation can be obtained using the bisection method as follows:

1. Choose two values of x, say a and b such that $f\left(a\right)f\left(b\right)<0$.

2. Estimate the root by $x_1=\frac{a+b}{2}$.

3. If, $f\left(a\right)f\left(x_1\right)<0$, the root would lie between a and $x_1$. Next, assume the next root to be $x_2=\frac{a+x_1}{2}$. If, $f\left(a\right)f\left(x_1\right)>0$, the root would lie between b and $x_1$. Now assume the next root to be $x_2=\frac{b+x_1}{2}$, and if $f\left(a\right)f\left(x_1\right)=0$, the root is $x_1$.

Calculation:

For the function $f\left(x\right)$:

$\begin{array}{c}f\left(-1\right)=-2.75{\left(-1\right)}^3+18{\left(-1\right)}^2-21\left(-1\right)-12\\ =29.75\\ f\left(0\right)=-2.75{\left(0\right)}^3+18{\left(0\right)}^2-21\left(0\right)-12\\ =-12\end{array}$

Thus,

$f\left(-1\right)f\left(0\right)<0$

Calculate the first root to be,

$\begin{array}{c}{x}_1=\frac{-1+0}{2}\\ =-0.5\end{array}$

Further,

$\begin{array}{c}f\left(-0.5\right)=-2.75{\left(-0.5\right)}^3+18{\left(-0.5\right)}^2-21\left(-0.5\right)-12\\ =3.34375\end{array}$

Thus, $f\left(-0.5\right)f\left(0\right)<0$. This implies that the root lies between $-0.5$ and 0.

Calculate the second root to be:

$\begin{array}{c}{x}_2=\frac{-0.5+0}{2}\\ =-0.25\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.25-\left(-0.5\right)}{-0.25}\right|\times 100\right)\%\\ =100\%\end{array}$

Therefore, the approximate relative percentage error is 100%.

Furthermore,

$\begin{array}{c}f\left(-0.25\right)=-2.75{\left(-0.25\right)}^3+18{\left(-0.25\right)}^2-21\left(-0.25\right)-12\\ =-5.58203\end{array}$

Thus, $f\left(-0.5\right)f\left(-0.25\right)<0$. This implies that the root lies between $-0.5$ and $-0.25$.

Calculate the third root to be:

$\begin{array}{c}{x}_3=\frac{-0.5+\left(-0.25\right)}{2}\\ =-0.375\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.375-\left(-0.25\right)}{-0.375}\right|\times 100\right)\%\\ =33.3\%\end{array}$

The approximate error is 33.3%.

Further,

$\begin{array}{c}f\left(-0.375\right)=-2.75{\left(-0.375\right)}^3+18{\left(-0.375\right)}^2-21\left(-0.375\right)-12\\ =-1.44873\end{array}$

Thus, $f\left(-0.5\right)f\left(-0.375\right)<0$. This implies that the root lies between $-0.5$ and $-0.375$.

Calculate the fourth root to be:

$\begin{array}{c}{x}_4=\frac{-0.5+\left(-0.375\right)}{2}\\ =-0.4375\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.4375-\left(-0.375\right)}{-0.4375}\right|\times 100\right)\%\\ =14.28\%\end{array}$

The approximate error is 14.28%.

Thus,

$\begin{array}{c}f\left(-0.4375\right)=-2.75{\left(-0.4375\right)}^3+18{\left(-0.4375\right)}^2-21\left(-0.4375\right)-12\\ =0.8631\end{array}$

Therefore, $f\left(-0.4375\right)f\left(-0.375\right)<0$. This implies that the root lies between $-0.4375$ and $-0.375$.

Calculate the fifth root to be:

$\begin{array}{c}{x}_5=\frac{-0.4375+\left(-0.375\right)}{2}\\ =-0.40625\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.40625-\left(-0.4375\right)}{-0.40625}\right|\times 100\right)\%\\ =7.69\%\end{array}$

The approximate error is 7.69%.

Thus,

$\begin{array}{c}f\left(-0.40625\right)=-2.75{\left(-0.40625\right)}^3+18{\left(-0.40625\right)}^2-21\left(-0.40625\right)-12\\ =-0.31367\end{array}$

Therefore, $f\left(-0.4375\right)f\left(-0.40625\right)<0$. This implies that the root lies between $-0.4375$ and $-0.40625$.

Calculate the sixth root to be:

$\begin{array}{c}{x}_6=\frac{-0.4375+\left(-0.40625\right)}{2}\\ =-0.42188\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.42188-\left(-0.40625\right)}{-0.42188}\right|\times 100\right)\%\\ =3.7\%\end{array}$

The approximate error is 3.7%.

Thus,

$\begin{array}{c}f\left(-0.42188\right)=-2.75{\left(-0.42188\right)}^3+18{\left(-0.42188\right)}^2-21\left(-0.42188\right)-12\\ =0.26947\end{array}$

Therefore, $f\left(-0.42188\right)f\left(-0.40625\right)<0$. This implies that the root lies between $-0.42188$ and $-0.40625$.

Calculate the seventh root to be:

$\begin{array}{c}{x}_7=\frac{-0.42188+\left(-0.40625\right)}{2}\\ =-0.41406\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.41406-\left(-0.42188\right)}{-0.41406}\right|\times 100\right)\%\\ =1.89\%\end{array}$

The approximate error is 1.89%.

Further,

$\begin{array}{c}f\left(-0.41406\right)=-2.75{\left(-0.41406\right)}^3+18{\left(-0.41406\right)}^2-21\left(-0.41406\right)-12\\ =-0.02341\end{array}$

Thus, $f\left(-0.42188\right)f\left(-0.41406\right)<0$. This implies that the root lies between $-0.42188$ and $-0.41406$.

Calculate the eighth root to be:

$\begin{array}{c}{x}_8=\frac{-0.42188+\left(-0.41406\right)}{2}\\ =-0.41797\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.41797-\left(-0.41406\right)}{-0.41797}\right|\times 100\right)\%\\ =0.93\%\end{array}$

The approximate error is 0.93%.

As, ${\epsilon }_a$ is less than 1%, the process is stopped and root can be approximated as $-0.41797$.

To determine

To calculate: The root of the equation $f\left(x\right)=-2.75x^3+18x^2-21x-12$ using the false-position method with $x_l$ and $x_u$ as $-1$ and 0, respectively, and with the stopping criteria being the iteration where ${\epsilon }_a$ falls below 1%.

Answer to Problem 4P

Solution: The root of the equation can be approximated as $-0.41402$.

Explanation of Solution

Given Information: The equation $f\left(x\right)=-2.75x^3+18x^2-21x-12$.

Initial guesses are,

$\begin{array}{l}{x}_l=-1\\ x_u=0\end{array}$

Formula used: A root of an equation can be obtained using the false-position method as follows:

1. Choose two values of x, say a and b such that $f\left(a\right)f\left(b\right)<0$.

2. Estimate the root by $x_1=b-\frac{f\left(b\right)\left(a-b\right)}{f\left(a\right)-f\left(b\right)}$.

3. If, $f\left(a\right)f\left(x_1\right)<0$, the root would lie between a and $x_1$. Next, assume the next root to be $x_2=x_1-\frac{f\left(x_1\right)\left(a-x_1\right)}{f\left(a\right)-f\left(x_1\right)}$. If, $f\left(a\right)f\left(x_1\right)>0$, the root would lie between $x_1$ and b. Now assume the next root to be $x_2=b-\frac{f\left(b\right)\left(x_1-b\right)}{f\left(x_i\right)-f\left(b\right)}$, and if $f\left(a\right)f\left(x_1\right)=0$, the root is $x_1$.

Calculation:

For the function $f\left(x\right)$:

$\begin{array}{c}f\left(-1\right)=-2.75{\left(-1\right)}^3+18{\left(-1\right)}^2-21\left(-1\right)-12\\ =29.75\\ f\left(0\right)=-2.75{\left(0\right)}^3+18{\left(0\right)}^2-21\left(0\right)-12\\ =-12\end{array}$

Thus,

$f\left(-1\right)f\left(0\right)<0$

Calculate the first root to be,

$\begin{array}{c}{x}_1=-1-\frac{\left(29.75\right)\left(-1-0\right)}{29.75-\left(-12\right)}\\ =-0.28743\end{array}$

Now,

$\begin{array}{c}f\left(-0.28743\right)=-2.75{\left(-0.28743\right)}^3+18{\left(-0.28743\right)}^2-21\left(-0.28743\right)-12\\ =-4.41173\end{array}$

Thus, $f\left(-1\right)f\left(-0.28743\right)<0$. This implies that the root lies between $-1$ and $-0.28743$.

Calculate the second root to be:

$\begin{array}{c}{x}_2=-1-\frac{\left(29.75\right)\left(-1-\left(-0.28743\right)\right)}{29.75-\left(-4.41173\right)}\\ =-0.37945\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.37945-\left(-0.28743\right)}{0.58802}\right|\times 100\right)\%\\ =24.25\%\end{array}$

The approximate relative percentage error is 24.25%.

Further,

$\begin{array}{c}f\left(-0.37945\right)=-2.75{\left(-0.37945\right)}^3+18{\left(-0.37945\right)}^2-21\left(-0.37945\right)-12\\ =-1.28966\end{array}$

Thus, $f\left(-1\right)f\left(-0.37945\right)<0$. This implies that the root lies between $-1$ and $-0.37945$.

Calculate the third root to be:

$\begin{array}{c}{x}_3=-1-\frac{\left(29.75\right)\left(-1-\left(-0.37945\right)\right)}{29.75-\left(-1.28966\right)}\\ =-0.40523\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.40523-\left(-0.37945\right)}{0.40523}\right|\times 100\right)\%\\ =6.36\%\end{array}$

The approximate error is 6.36%.

Now,

$\begin{array}{c}f\left(-0.40523\right)=-2.75{\left(-0.40523\right)}^3+18{\left(-0.40523\right)}^2-21\left(-0.40523\right)-12\\ =-0.35129\end{array}$

Thus, $f\left(-1\right)f\left(-0.40529\right)<0$. This implies that the root lies between $-1$ and $-0.40529$.

Calculate the fourth root to be:

$\begin{array}{c}{x}_4=-1-\frac{\left(29.75\right)\left(-1-\left(-0.40529\right)\right)}{29.75-\left(-0.35129\right)}\\ =-0.41217\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.41217-\left(-0.40529\right)}{-0.41217}\right|\times 100\right)\%\\ =1.68\%\end{array}$

The approximate error is 1.68%.

Now,

$\begin{array}{c}f\left(-0.41217\right)=-2.75{\left(-0.41217\right)}^3+18{\left(-0.41217\right)}^2-21\left(-0.41217\right)-12\\ =-0.09384\end{array}$

Thus, $f\left(-1\right)f\left(-0.41217\right)<0$. This implies that the root lies between $-1$ and $-0.41217$.

Calculate the fifth root would be:

$\begin{array}{c}{x}_5=-1-\frac{\left(29.75\right)\left(-1-\left(-0.41217\right)\right)}{29.75-\left(-0.09384\right)}\\ =-0.41402\end{array}$

The approximate error is computed as:

$\begin{array}{c}{\epsilon }_a=\left(\left|\frac{-0.41402-\left(-0.41217\right)}{-0.41217}\right|\times 100\right)\%\\ =0.45\%\end{array}$

The approximate error is 0.45%.

As, ${\epsilon }_a$ is less than 1%, the process is stopped and root can be approximated as $-0.41402$.