c > 1.If f(x) = 2

^{x}, which of these 5 functions yields the greatest value for f(g(x)), for all

x > 1?

B. g(x) = c/x

C. g(x) = x/c

D. g(x) = x – c

E. g(x) = log

_{c}x

Listed below are 5 functions, each denoted g(x) and each involving a real number constant

c > 1.If f(x) = 2^{x}, which of these 5 functions yields the greatest value for f(g(x)), for all

x > 1?

c > 1.If f(x) = 2

x > 1?

A. g(x) = cx

B. g(x) = c/x

C. g(x) = x/c

D. g(x) = x – c

E. g(x) = log_{c}x

B. g(x) = c/x

C. g(x) = x/c

D. g(x) = x – c

E. g(x) = log

Tutors, sign in to answer this question.

John M. | John - Algebra TutorJohn - Algebra Tutor

f(x)=2^{x}

Then f(g(x))= 2^{g(x)}

we are supposed evalute the 5 expressions to determine which would yield the highest value of f(x). Because the g(x) functions share a common base, we can evalute which one would have the highest value, and that would yeild the highest value for f(x).^{
}

For any value of c, answer A. is clearly larger than a fraction (B. and C.), and larger than D. The only sticky point is answer E, is cx larger than log_{c}x?

-------

In general the expression: log_{a}b=n can be written as a^{n}=b

For example: let a = 2 and b=8. We know that n=3 because 8 is 2^{3}. Similarly, log_{2}8=3.

------

returning to the problem, we know that the value of log_{c}x must be smaller than x for any c, therefore it is less than "cx" as well.

So the answer is: A. g(x)=cx, yeilds the highest value of f(g(x)) for all x>1.

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments