Question

question 11
what is the slope of the tangent line to $f(x) = \frac{1}{x^2}$ at $x = 2$?
$\circ \\ \frac{1}{4}$
$\circ \\ -\frac{1}{4}$
$\circ \\ \frac{1}{2}$
$\circ \\ -\frac{1}{2}$
question 12

Explanation:

Step1: Rewrite the function

Rewrite \( f(x)=\frac{1}{x^2} \) as \( f(x) = x^{-2} \).

Step2: Find the derivative

Use the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \). For \( f(x)=x^{-2} \), the derivative \( f^\prime(x)=-2x^{-3}=-\frac{2}{x^3} \).

Step3: Evaluate the derivative at \( x = 2 \)

Substitute \( x = 2 \) into \( f^\prime(x) \): \( f^\prime(2)=-\frac{2}{2^3}=-\frac{2}{8}=-\frac{1}{4} \).

Answer:

\(-\frac{1}{4}\) (corresponding to the option: \(-\frac{1}{4}\))