# Determine the end behavior of each function...?

• September 8th 2010, 07:27 PM
melliep
Determine the end behavior of each function...?
a) f(x) = 17 + 3x^2 - 13x^3 - 6x^4

b) f(x) = (5x^2-7x+2)/(4x^2-2)

c) e^x

all of them are asking for two answers. the first as x to -infinity and the second as x to infinity.

thank you thank you =)
• September 8th 2010, 07:37 PM
undefined
Quote:

Originally Posted by melliep
a) f(x) = 17 + 3x^2 - 13x^3 - 6x^4

b) f(x) = (5x^2-7x+2)/(4x^2-2)

c) e^x

all of them are asking for two answers. the first as x to -infinity and the second as x to infinity.

thank you thank you =)

Have you done these kinds of limits before?

a) Which term dominates the others as x approaches +/- infinity?

b) Which term in denominator dominates, which term in numerator..?

c) If it's not obvious.. which I guess it's not since you're asking.. just graph it. Then justify it.
• September 8th 2010, 07:42 PM
mr fantastic
Quote:

Originally Posted by melliep
a) f(x) = 17 + 3x^2 - 13x^3 - 6x^4

b) f(x) = (5x^2-7x+2)/(4x^2-2)

c) e^x

all of them are asking for two answers. the first as x to -infinity and the second as x to infinity.

thank you thank you =)

b) You could also do a polynomial long division to get $\displaystyle f(x) = \frac{-7x+\frac{9}{2}}{4x^2 - 2} + \frac{5}{4}$ ....
• September 8th 2010, 08:04 PM
melliep
mmm I'm really sorry that doesn't help me.
I've never learned this before and I'm expected to since I'm in adv. calc.
• September 8th 2010, 08:05 PM
undefined
Yet another way to handle (b) is

(x won't be 0)

$\displaystyle \frac{5x^2-7x+2}{4x^2-2}=\frac{\frac{1}{x^2}(5x^2-7x+2)}{\frac{1}{x^2}(4x^2-2)}=\frac{5-\frac{7}{x}+\frac{2}{x^2}}{4-\frac{2}{x^2}}$
• September 8th 2010, 08:13 PM
undefined
Quote:

Originally Posted by melliep
mmm I'm really sorry that doesn't help me.
I've never learned this before and I'm expected to since I'm in adv. calc.

This might help, I found it w/ google search and seems good at a glance

Limits to Infinity
• September 9th 2010, 03:00 AM
mr fantastic
Quote:

Originally Posted by melliep
mmm I'm really sorry that doesn't help me.
I've never learned this before and I'm expected to since I'm in adv. calc.

Sorry to ask, but how did you get into an advanced calculus class without having the necessary mathematical pre-requisites?
• September 15th 2010, 10:40 PM
melliep
Hahaha...thanks...
Mmm because I'm smart, maybe?
And because I'm good at taking tests.
Sorry if my question was not up to par.