Which of the following function's output decreases without bound as $x$ increases without bound? $f(x)=x^{3}-118$ $f(x)=1-x^{2}+x^{3}$ $f(x)=x^{7}-x^{5}-x^{3}-12$ $f(x)=1-x^{2}$ $f(x)=x^{2}-12 x-27$

Step 1 ：The question is asking which function's output decreases without bound as $x$ increases without bound. This means we are looking for a function that will continue to produce smaller and smaller values as $x$ gets larger and larger.

Step 2 ：To determine this, we need to look at the highest degree term in each function. This is because as $x$ gets larger, the highest degree term will dominate the function's behavior.

Step 3 ：If the highest degree term is positive and has an even degree, the function will increase without bound as $x$ increases. If the highest degree term is positive and has an odd degree, the function will decrease without bound as $x$ decreases and increase without bound as $x$ increases. If the highest degree term is negative and has an even degree, the function will decrease without bound as $x$ increases. If the highest degree term is negative and has an odd degree, the function will increase without bound as $x$ decreases and decrease without bound as $x$ increases.

Step 4 ：Looking at the functions, we can see that the highest degree term in each function is as follows: $f(x)=x^{3}-118$ has a highest degree term of $x^{3}$ which is positive and odd. $f(x)=1-x^{2}+x^{3}$ has a highest degree term of $x^{3}$ which is positive and odd. $f(x)=x^{7}-x^{5}-x^{3}-12$ has a highest degree term of $x^{7}$ which is positive and odd. $f(x)=1-x^{2}$ has a highest degree term of $-x^{2}$ which is negative and even. $f(x)=x^{2}-12 x-27$ has a highest degree term of $x^{2}$ which is positive and even.

Step 5 ：From our earlier thought, we know that a function will decrease without bound as $x$ increases if the highest degree term is negative and even. Therefore, the function that decreases without bound as $x$ increases is $f(x)=1-x^{2}$.

Step 6 ：Final Answer: The function that decreases without bound as $x$ increases is $f(x)=1-x^{2}$. Therefore, the final answer is \(\boxed{f(x)=1-x^{2}}\).