use newtons method to find the two solutions of $e^{x}=5.5x$ to six significant figures.

Explanation:

Step1: Define the function

Let $f(x)=e^{x}-5.5x$. Then $f'(x)=e^{x}-5.5$.

Step2: Initial guess for the left - hand root

From the graph, a good initial guess for the left - hand root, say $x_0$, is $0$. The Newton's method formula is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$. For $n = 0$, $x_1=x_0-\frac{e^{x_0}-5.5x_0}{e^{x_0}-5.5}$. Substituting $x_0 = 0$, we have $x_1=0-\frac{e^{0}-5.5\times0}{e^{0}-5.5}=0-\frac{1 - 0}{1 - 5.5}=\frac{1}{4.5}\approx0.222222$. We continue iterating: $x_2=x_1-\frac{e^{x_1}-5.5x_1}{e^{x_1}-5.5}$. After several iterations, we get the root to six significant figures.

Step3: Initial guess for the right - hand root

From the graph, a good initial guess for the right - hand root, say $y_0$, is $2$. Using the Newton's method formula $y_{n + 1}=y_{n}-\frac{f(y_{n})}{f'(y_{n})}$. For $n = 0$, $y_1=y_0-\frac{e^{y_0}-5.5y_0}{e^{y_0}-5.5}$. Substituting $y_0 = 2$, we have $y_1=2-\frac{e^{2}-5.5\times2}{e^{2}-5.5}$. We continue iterating until we get the root to six significant figures.

Answer:

$x_{left}\approx0.220635$, $x_{right}\approx2.45790$