Answer:
1. 3/5
2. 1.219841
3. 2 - f(2)/7
4. -1.9682
5. 1.502446
Step-by-step explanation:
If we call
then the second approximation to the root of the equation f(x)=0 would be
hence
2.
Here we have
Let's start with
then
Since the first 6 decimals of and are equal, the desired approximation is 1.219841
3.
If the line y = 4x − 1 is tangent to the curve y = f(x) when x = 2, then f'(2) = 4*2 - 1 = 7, so
4.
Here
5.
We want to find all the values x such that
or what is the same, the x such that
so, let f(x) be
and let's use Newton's method to find the roots of f(x).
Since
f is strictly increasing, and since
f(1) = e+3-9 = e - 6 < 0
and
f has only one root in [1,2]
By using Newton's iterations starting with
Since then
x=1.50244564 is the desired root.
The answers as a comma-separated list would be
1, 1.573899431, 1.503982961, 1.502446348, 1.50244564
The answer rounded to six decimal places would be
1.502446
