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Mathematics 7 Online
OpenStudy (anonymous):

Algebra 2 please help

OpenStudy (anonymous):

For what values of b is f(x) = b^x an increasing function?

OpenStudy (anonymous):

A. Any real number B. b > 0 C. b > 1 D. 0< b <1

OpenStudy (anonymous):

@pgpilot326

OpenStudy (anonymous):

@shamil98

OpenStudy (anonymous):

these are exponentials and b>0 (otherwise if b<0 it's a complex valued function). So you can ignore the first choice. you should see that one of them works and the others don't. try a couple... start with a simple one... say the last is it increasing?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

let's look... 0<1/2<1/ (1/2)^1 = 1/2 (1/2)^2 = 1/4 so as x increases, f(x) decreases... not an increasing function.

OpenStudy (anonymous):

so it would be D right

OpenStudy (anonymous):

what if b = 1. then 1^1 = 1 and 1^2 = 1 so b = 1 is neither increasing nor decreasing. (it's the same for all of these if you use negative exponents, or fractional exponents or real exponents... so long as the exponents are increasing)

OpenStudy (anonymous):

no. D is not right.

OpenStudy (anonymous):

because at first i thought it was B but then you said is neither increasing nor decreasing so yeah

OpenStudy (anonymous):

look at the values in those intervals... it would have to be true for all values of b in the interval. the only one that works is C.

OpenStudy (anonymous):

D is strictly decreasing. C is strictly increasing B is increasing, decreasing and neither depending on the value (0<b<1 it's decreasing, b>1 it's increasing, b = 1 it's neither) for A, it's increasing, decreasing, neither and a complex valued function depending on the value of b. (0<b<1 it's decreasing, b>1 it's increasing, b = 1 or 0 it's neither, if b<0 it's a complex valued function)

OpenStudy (anonymous):

did i help you?

OpenStudy (anonymous):

yes a lot thank you very much if i could give you more medals i would because you really did help thank you so much :) <3

OpenStudy (anonymous):

you're welcome

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