Question

Consider the following graph of y = f(x):

Which of the following could be the equation for f(x)?

I. Log3(ax^2), where a < 0.
II. Log5(ax^5), where a > 0.
III. Log6 (-(ax)^3), where a > 0.
(Hint: Simplify them algebraically first.)


A) I only
B) II only
C) III only
D) Both I and III
E) None of the above

Answer 1

this is the graph:

Answer 2

oh...ok....lol your second guessing yourself.

Answer 3

anyway thanks!

Answer 4

As third guessing myself, I am now more or less sure that 3 is the only one. This is its real-valued plot for a =2: http://www3.wolframalpha.com/Calculate/MSP/MSP17911a159fif90684f35000050c2934c4b7ibd68?MSPStoreType=image/gif&s=16&w=299&h=131&cdf=RangeControl
while 2, for a = 2 also: http://www4b.wolframalpha.com/Calculate/MSP/MSP9311a159hbhe6631h2b000025fh42i4f8cbi61h?MSPStoreType=image/gif&s=64&w=299&h=131&cdf=RangeControl
and I is undefined in the real, I suspect.

Answer 5

we're about to find out. ARE YOU READY!?

Answer 6

It was C!

Answer 7

Third time's the charm, right?

Answer 8

none of that could be if the domain is real

Answer 9

wow that took a battle.....thanks! :)

Answer 10

log must have input positive

Answer 11

so if domain is real,the input will be negative

Answer 12

you can take as a function just if you restrict you domain

Answer 13

your*

Answer 14

lol, I submitted the answer and got it right.(online quiz) so...if it's wrong I don't care now. If it is wrong, then the quiz makers made an error.

Answer 15

Exactly @RaphaelFilgueiras, as it's x^3, x must be negative in order to the log to be defined. So, that is why the graph goes into the negative x direction.

Answer 16

Like I said in my first post, -(ax)^3 will only be undefined if (ax)^3 is positive or zero. If x is negative, as a is positive, the product will be negative, a negative number raised to an odd exponent is negative times negative one produces a positive answer.