Answer:
the least integer for n is 2
Step-by-step explanation:
We are given;
f(x) = ln(1+x)
centered at x=0
Pn(0.2)
Error < 0.01
We will use the format;
[[Max(f^(n+1) (c))]/(n + 1)!] × 0.2^(n+1) < 0.01
So;
f(x) = ln(1+x)
First derivative: f'(x) = 1/(x + 1) < 0! = 1
2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1
3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2
4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6
This follows that;
Max|f^(n+1) (c)| < n!
Thus, error is;
(n!/(n + 1)!) × 0.2^(n + 1) < 0.01
This gives;
(1/(n + 1)) × 0.2^(n + 1) < 0.01
Let's try n = 1
(1/(1 + 1)) × 0.2^(1 + 1) = 0.02
This is greater than 0.01 and so it will not work.
Let's try n = 2
(1/(2 + 1)) × 0.2^(2 + 1) = 0.00267
This is less than 0.01.
So,the least integer for n is 2
y is -2 is the answer for the question
Answer:
Step-by-step explanation: when you first open the graphing tool you have to click relationship and choose custom. put in gwen’s equations (y=100+10x) then go to relationship and choose custom again and put in tristan’s equation (y=12.5)
also click on the settings button at the bottom of the graph and put the x axis to -100 min and 100 max and for y axis put -1000 min and 1000 max or you could choose your own numbers but thats just what i put
Hello
A. will be your answer.
Hoped This Helped! :D
I = PRT
21 = 200 * 0.035 * T
21 = 7T
3 years = T
