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Mathematics 20 Online
OpenStudy (anonymous):

Which functions in the table below give values that could come from exponential functions?

OpenStudy (anonymous):

OpenStudy (aum):

x increases by 1 each time. See how f(x) increases. Take the difference in f(x) values: 3, 5, 7, 9, .... Take the difference again: 2, 2, 2, ... So the second differences of f(x) are constant. This implies f(x) is quadratic and not exponential

OpenStudy (aum):

See how g(x) increases. It is doubling every time. 0.125 x 2 = 0.25; 0.25 x 2 = 0.5; 0.5 x 2 = 1, ... This is a geometric sequence. Therefore it is an exponential function.

OpenStudy (aum):

Can you do the last two?

OpenStudy (anonymous):

ohhh so is h(x) is one right?

OpenStudy (aum):

h(x) increases by the SAME amount every time. It increases by 0.25. So it is an arithmetic sequence which is linear or a straight line. So h(x) is NOT exponential. Try the last one.

OpenStudy (anonymous):

but doesn't k(x) go down?

OpenStudy (aum):

Yes, it does. Exponential GROWTH function will go up but exponential DECAY function will go down. So just because it goes down you can't say it is not exponential.

OpenStudy (aum):

Observe the pattern and see what is happening to the k(x) values each time x increases by 1.

OpenStudy (aum):

How to get from 64 to 16? From 16 to 4? Is there a pattern?

OpenStudy (anonymous):

Yea...but idk what

OpenStudy (aum):

How to go from 64 to 16? You divide by 4. How to go from 16 to 4? You divide by 4. So the pattern is you divide by 4 each time. This is a geometric sequence with a common ratio of 1/4. Therefore, it is an exponential function. (BTW, there is a mistake in the table they gave you. When x = 0, k(x) must be 1 and not 0.)

OpenStudy (anonymous):

ohhhhhh wow thank so much for explaining it out like that to me. Do you think you can help me with another question?

OpenStudy (aum):

Close this one post a new one and tag me. In the meantime there are others who have tagged me and I will come to you after that.

OpenStudy (anonymous):

ok thankyou :)

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