SL 2.2—Functions, domains, range, graphs

The function is $E(L)=\frac{600}{L-10}$ with domain $L>10$. We need to find all possible output values.

Since $L>10$, the denominator $(L-10)>0$. The numerator $600$ is a positive constant, so $E(L)=\frac{600}{L-10}>0$ for all valid inputs.

As $L\to 10^{+}$, the denominator approaches $0^{+}$, so $E(L)\to +\infty$. As $L\to \infty$, $E(L)\to 0^{+}$. So $E$ can get arbitrarily close to $0$ but never equals $0$.

Since $E$ takes all positive values but never reaches $0$ or any fixed lower bound above $0$, the range is $E>0$.

We need the range of $E(L)=\frac{600}{L-10}$ for $L>10$, meaning all possible output values.

The option $E\ge 0$ includes $E=0$, but $\frac{600}{L-10}$ can never equal $0$ since the numerator is $600\ne 0$. Eliminated.

The option $E>10$ would exclude values like $E=5$, but substituting $L=130$ gives $E=\frac{600}{120}=5$, which is valid. Eliminated.

The option $E\ge 60$ would require a minimum value of $60$. But for large $L$, $E$ approaches $0^{+}$, so values below $60$ are achievable. Eliminated.

The only remaining option, $E>0$, correctly captures that the output is always strictly positive but unbounded above and approaches $0$ from above.

The domain is $-1\le x\le 3$, so we evaluate $f$ at the endpoints $x=-1$ and $x=3$.

$f(-1)=3(-1{)}^3-4=3(-1)-4=-3-4=-7$

$f(3)=3(3{)}^3-4=3(27)-4=81-4=77$

Since $f(x)=3x^3-4$ is a strictly increasing cubic (its derivative $f^{\prime }(x)=9x^2\ge 0$ everywhere, equal to zero only at $x=0$), there are no local extrema that would affect the range on this interval.

The function increases from $-7$ to $77$ over the given domain, so the range is $-7\le f(x)\le 77$.

We need the range of $f(x)=3x^3-4$ on $[-1,3]$, i.e., the minimum and maximum output values.

$f(-1)=3(-1)-4=-7$ and $f(3)=3(27)-4=77$. The minimum is $-7$ and maximum is $77$.

This option uses $-4$ as the minimum, which is $f(0)$, not the minimum on $[-1,3]$. Since $f(-1)=-7<-4$, this lower bound is incorrect. Eliminated.

This uses $81$ as the upper bound, which equals $3(27)=81$ before subtracting $4$. The correct value is $81-4=77$, not $81$. Eliminated.

With minimum $-7$ and maximum $77$, both endpoints included, the range is $-7\le f(x)\le 77$.