## University Calculus: Early Transcendentals (3rd Edition)

$F(x)$ has an absolute maximum value of $-0.25$ at $x=2$ and an absolute minimum value of $-4$ at $x=0.5$ on the defined domain.
$$F(x)=-\frac{1}{x^2}\hspace{1cm}0.5\le x\le2$$ 1) Find the derivative $F'(x)$: $$F'(x)=-\frac{(1)'\times x^2-1\times(x^2)'}{x^4}=-\frac{0\times x^2-2x}{x^4}$$ $$F'(x)=-\frac{(-2x)}{x^4}=\frac{2}{x^3}$$ Here there is no value of $x$ for which $F'(x)=0$. However, we can see that $F'(x)$ is undefined at $x=0$, which means $x=0$ is a critical point of $F(x)$. However, since the defined domain here is $[0.5,2]$, we would not take $x=0$ into consideration. 2) Evaluate $F(x)$ at the endpoints: - For $x=0.5$: $$F(0.5)=-\frac{1}{0.5^2}=-\frac{1}{0.25}=-4$$ - For $x=2$: $$F(2)=-\frac{1}{2^2}=-\frac{1}{4}=-0.25$$ This means the function $F(x)$ has an absolute maximum value of $-0.25$ at $x=2$ and an absolute minimum value of $-4$ at $x=0.5$ on the defined domain.