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Given the function \( f(x) = \frac{1}{x^2} \), find the value of \( f(x) \) at \( x = 2 \) and determine the slope of the tangent line to the curve at that point.
To find the value of \( f(x) \) at \( x = 2 \) and the slope of the tangent line to the curve at that point for the function \( f(x) = \frac{1}{x^2} \), we will follow these steps:
1、Evaluate \( f(x) \) at \( x = 2 \):
\[
f(2) = \frac{1}{2^2} = \frac{1}{4}
\]
So, the value of \( f(x) \) at \( x = 2 \) is \( \frac{1}{4} \).
2、Find the derivative \( f'(x) \):
The function \( f(x) = \frac{1}{x^2} \) can be rewritten as \( f(x) = x^{-2} \)、Using the power rule for differentiation, which states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \), we get:
\[
f'(x) = -2x^{-3} = -\frac{2}{x^3}
\]
So, the derivative of \( f(x) \) is \( f'(x) = -\frac{2}{x^3} \).
3、Evaluate the derivative at \( x = 2 \):
\[
f'(2) = -\frac{2}{2^3} = -\frac{2}{8} = -\frac{1}{4}
\]
So, the slope of the tangent line to the curve at \( x = 2 \) is \( -\frac{1}{4} \).
Therefore, the value of \( f(x) \) at \( x = 2 \) is \( \frac{1}{4} \) and the slope of the tangent line to the curve at that point is \( -\frac{1}{4} \)、The final answer is:
\[
\boxed{-\frac{1}{4}}
\]