If f^(')(x)=[f(x)]^(2) and f(0)=1, then f(6)=1//n for some integer n. What is n ? n=

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If  f^(')(x)=[f(x)]^(2) and  f(0)=1, then  f(6)=1//n for some integer  n. What is  n ?  n=

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Find f'(0): Since f'(x) = [f(x)]^2 and f(0) = 1, we can start by finding f'(0) using the given information. f'(0) = [f(0)]^2 = 1^2 = 1.Rate of Change at x=0: The derivative f'(x) represents the rate of change of f(x). Since f'(0) = 1, the function f(x) is increasing at x = 0.Behavior of f(x): We know that f(x) is increasing and f(0) = 1. If f'(x) = [f(x)]^2, then f(x) must remain positive and increasing for all x > 0.Integrate to find f(6): To find f(6), we need to integrate f'(x) from 0 to 6. However, we are given that f(6) = 1/n, so we need to find n that satisfies this condition.Understanding f'(x): Since we don't have an explicit function for f(x), we cannot directly integrate f'(x). But we can use the fact that f'(x) = [f(x)]^2 to understand the behavior of f(x).Implications of f'(x): If f'(x) = [f(x)]^2, then as x increases, f(x) will increase at a rate proportional to the square of f(x). This implies that f(x) will grow very quickly.Determining n: Given that f(6) = 1/n, and knowing that f(x) grows quickly, n must be a very large number because f(6) is a very small positive number.However, without an explicit form for f(x), we cannot find the exact value of n. We need more information or a different approach to solve for n.

If $f^{\prime}(x)=[f(x)]^{2}$ and $f(0)=1$ , then $f(6)=1 / n$ for some integer $n$ . $\newline$ What is $n$ ? $\newline$ $n=$

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If f′(x)=[f(x)]2 f^{\prime}(x)=[f(x)]^{2} f′(x)=[f(x)]2 and f(0)=1 f(0)=1 f(0)=1, then f(6)=1/n f(6)=1 / n f(6)=1/n for some integer n n n.\newlineWhat is n n n ?\newlinen= n= n=

If $f^{\prime}(x)=[f(x)]^{2}$ and $f(0)=1$ , then $f(6)=1 / n$ for some integer $n$ . $\newline$ What is $n$ ? $\newline$ $n=$