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
Answer:
f(g(5)) = 64
g(f(5)) = 28
Step-by-step explanation:
Given that f(x) = x^2 and g(x) = x+3
f(g(x) = f(x+3)
f(x+3) = (x+3)^2
f(g(x)) = (x+3)^2
f(g(5)) = (5+3)^2
f(g(5)) = 8^2
f(g(5)) = 64
b) g(f(x)) = g(x^2)
g(f(x)) = x^2 + 3
g(f(5)) = 5^2 +3
g(f(5)) = 25 + 3
g(f(5)) = 28
Hence the value of g(f(5)) is 28
Answer:
1
Step-by-step explanation:
One is the smallest multiple of 50. We can find this by finding all the multiples of 50. The multiples of 50 are: 1, 2, 5, 10, 25, and 50.
We can see that one is the smallest multiple because it is less than 2, 5, 10, 25 and 50. So, that means the answer to the question is 1.
Hope this helps.
Answer:
13
Step-by-step explanation:
17-4=13 therefore 17-13=4
Answer:
260 / 5 = 52
Step-by-step explanation:
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