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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Transformations of quadratic functions

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Q. 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=

Find $f'(0)$ : Since $f'(x) = [f(x)]^2$ and $f(0) = 1$ , we can start by finding $f'(0)$ using the given information. $\newline$ $f'(0) = [f(0)]^2 = 1^2 = 1$ .
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 $f(6) = \frac{1}{n}$ , 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 $f(6) = \frac{1}{n}$ , and knowing that $f(x)$ grows quickly, $n$ must be a very large number because $f(6)$ is a very small positive number.
Determining $n$ : Given that $f(6) = \frac{1}{n}$ , 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 $n$ .