Question

ex-6 Step 1 of 4 Recall that Newton's method is used to approximate a zero of a function, f(x), using the equation below, where $x_0$ is an initial guess. $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ To determine where $e^x - 6 = x$, we find the zeros of the function $f(x) = e^x - x - 6$. We will also need the derivative function, $f'(x) = e^x - 1$ Observe from a graph of the function that there is one positive and one negative root. We will start with finding the positive root. For the initial guess $x_0$, we will use the nearest positive integer to the exact root, that is, $x_0 = 2$ Step 2 of 4 We apply Newton's Method until we have the same value in successive iterations, up to five decimal places. If $x_0 = 2$, then $f(2) = e^2 - 8$ and $f'(2) = e^2 - 1$, so we calculate $x_1$ as follows. (Round your answer for $x_1$ to seven decimal places.) $x_1 = 2 - \frac{f(2)}{f'(2)}$ = 2.0956235 2.0956235 Step 3 of 4 Using $x_1 = 2.0956235$, we calculate the next few iterations. (Round your answers to seven decimal places.) $x_2 = 2.0956235 - \frac{f(2.0956235)}{f'(2.0956235)}$ = 2.1912469 $x_3 = $ $x_4 = $

          ex-6
Step 1 of 4
Recall that Newton's method is used to approximate a zero of a function, f(x), using the equation below, where $x_0$ is an initial guess.
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
To determine where $e^x - 6 = x$, we find the zeros of the function $f(x) = e^x - x - 6$.
We will also need the derivative function, $f'(x) = e^x - 1$
Observe from a graph of the function that there is one positive and one negative root.
We will start with finding the positive root.
For the initial guess $x_0$, we will use the nearest positive integer to the exact root, that is, $x_0 = 2$
Step 2 of 4
We apply Newton's Method until we have the same value in successive iterations, up to five decimal places.
If $x_0 = 2$, then $f(2) = e^2 - 8$ and $f'(2) = e^2 - 1$, so we calculate $x_1$ as follows. (Round your answer for $x_1$ to seven decimal places.)
$x_1 = 2 - \frac{f(2)}{f'(2)}$
= 2.0956235
2.0956235
Step 3 of 4
Using $x_1 = 2.0956235$, we calculate the next few iterations. (Round your answers to seven decimal places.)
$x_2 = 2.0956235 - \frac{f(2.0956235)}{f'(2.0956235)}$
= 2.1912469
$x_3 = $
$x_4 = $

ex-6
Step 1 of 4
Recall that Newton's method is used to approximate a zero of a function, f(x), using the equation below, where x0 is an initial guess.
xn+1 = xn - (f(xn))/(f'(xn))
To determine where e^x - 6 = x, we find the zeros of the function f(x) = e^x - x - 6.
We will also need the derivative function, f'(x) = e^x - 1
Observe from a graph of the function that there is one positive and one negative root.
We will start with finding the positive root.
For the initial guess x0, we will use the nearest positive integer to the exact root, that is, x0 = 2
Step 2 of 4
We apply Newton's Method until we have the same value in successive iterations, up to five decimal places.
If x0 = 2, then f(2) = e^2 - 8 and f'(2) = e^2 - 1, so we calculate x1 as follows. (Round your answer for x1 to seven decimal places.)
x1 = 2 - (f(2))/(f'(2))
= 2.0956235
2.0956235
Step 3 of 4
Using x1 = 2.0956235, we calculate the next few iterations. (Round your answers to seven decimal places.)
x2 = 2.0956235 - (f(2.0956235))/(f'(2.0956235))
= 2.1912469
x3 =
x4 =

Added by Theresa H.

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