Question

The end behavior of f(x) = 9+2x^3 / x^4-81
most closely matches which of the following?

y = 0
y = 2
y = 9
y = −9

Answer 1

Evaluate
f(10)
f(100)
f(1000)

(use a calculator) and tell me when happens to the function as x grows.

Answer 2

okay gimme a sec

Answer 3

f(10)= .2025

Answer 4

f(100)= .02000

Answer 5

it gets smaller @SolomonZelman

Answer 6

Yes, and it would continue getting smaller and smaller....

Answer 7

it actually goes to zero...

Answer 8

Do you want to leave with that, or do you want to verify that this is so?

Answer 9

alright so zero is the answer. I've got it thanks ;)

Answer 10

For very large values of x (like x=1000 and bigger)
\(\large\color{#000000 }{ \displaystyle 2x^3>9>0 }\)
and (also for very large x-values); \(\large\color{#000000 }{ \displaystyle \frac{1}{2}x^4<x^4-81<x^4 }\)

And the following would be also true:
\(\large\color{#000000 }{ \displaystyle \frac{2x^3}{x^4}<\frac{9+2x^3}{x^4-81} <\frac{2x^3+2x^3}{x^4-\frac{1}{2}x^4} }\)

Answer 11

Do you agree with the last statement?

(left side => smaller numerator, bigger denominator
right side => bigger numerator, smaller denominator

so, left < middle < right)

Answer 12

What will this do?
We can simplify both right and left sides and show that they surely go to zero, and therefore the middle (our function would also go to 0.

Answer 13

\(\large\color{#000000 }{ \displaystyle \frac{2x^3}{x^4}<\frac{9+2x^3}{x^4-81} <\frac{2x^3+2x^3}{x^4-\frac{1}{2}x^4} }\)

\(\large\color{#000000 }{ \displaystyle \frac{2}{x}<\frac{9+2x^3}{x^4-81} <\frac{24x^3}{\frac{1}{2}x^4} }\)

\(\large\color{#000000 }{ \displaystyle \frac{2}{x}<\frac{9+2x^3}{x^4-81} <\frac{48}{x} }\)

Answer 14

and left and right surely go to zero.