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
8.8 squared + 9.5 squared = square root 12.94 rounded 12.1
The answer:
by definition, an exponential function with base c is defined by <span>h (x) = ac^x</span><span>
where a ≠0, c > 0 , b ≠1, and x is any real number.</span>
The base, c, is a constant and the exponent, x<span>, is a variable.
</span>so if we have f(x)=3(3\8)^2x, this equivalent to f(x)=3(3\8)^y(x),
where y (x)=2x, <span>
therefore, the base is 3/8, and the variable is the function </span>y (x)=2x,
Answer:
Graph 3 on Edg.
Step-by-step explanation:
