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f (x) = x^2 + 2^x for all integers x. f (k)
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01 Dec 2023, 19:05
When the comparison column is 0, I find that it is a good tactic to consider -1 and 1 as inputs for k, as well as when k = 0.
\(f(-1) = \frac{3}{2}\)
\(f(0) = 1\)
\(f(1) = 3\)
f(-1) gets us exactly one of the solutions. However, we can see there is also some value of k between 0 and 1 that would get us \(\frac{3}{2}\). (The exact value of this k isn't necessary to answer this question, just that it's greater than 0.)
Therefore, there are 2 solutions for k on both sides of 0, so the answer is D.