If f is the function defined for all k such that f(k) = k^5/16, what is f(2k) in terms of f(k)?
a. 1/8 f(k)
b. 5/8 f(k)
c. 2 f(k)
d. 10 f(k)
e. 32 f(k)
a. 1/8 f(k)
b. 5/8 f(k)
c. 2 f(k)
d. 10 f(k)
e. 32 f(k)
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You are right... it should be 32/16 * f(k) = 2f(k).sl750 wrote:f(k)=k^5/16.
f(2k)=(2k)^5/16
Shouldn't it be (2^5/16)*(k^5/16)
(32)^1/16*f(k)
No I think the question is k raised to the power 5 and then this whole value divided by 16.sl750 wrote:More like 16th root of 32 *f(k)
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