Bunuel wrote:

For which of the following functions is f(−1/2) > f(2)?

A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2

Of the 5 answer options, you can straight away eliminate Options A, B, and C because all of these

options have an x or x^2 in the numerator. Substituting -\(\frac{1}{2}\) can never be greater than substituting 2.

Now that we are down to two answer options,

we can substitute the values to check in which case, value of f(-\(\frac{1}{2}\)) > value of f(\(2\))

In Option D, f(-\(\frac{1}{2}\)) = 3 + (-2) = 1 (because \(\frac{1}{\frac{-1}{2}}\) = -2) but f(\(2\)) = \(3 + \frac{1}{2} = \frac{7}{2}\). Here, f(-\(\frac{1}{2}\)) < f(\(2\))

Therefore, the only answer option we are left with is f(x) = \(\frac{3}{x^2}\)

(Option E) which is our answer!

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