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Find the maximum value of f(x). n is a constant

f(x)= C(x,2) *$2^{n-x}$
asked in Calculus | 178 views
0
$\frac{3}{8} * 2^{n}$ ?
0
yes. can you share you solution. ?I am doing some mistake and not able to find:(
+1
Didn't do by naive method coz the equation turns out to be complex.

Simply try for different values of x.

x cannot be less than 2, it is at least 2

For x = 2, f(x) = $\binom{2}{2}*2^{n-2}$ = $\frac{2^{n}}{4}$

For x = 3, f(x) = $\binom{3}{2}*2^{n-3}$ = $\frac{3}{8}*2^{n}$

For x = 4, f(x) = $\binom{4}{2}*2^{n-4}$ = $\frac{3}{8}*2^{n}$

For x = 5, f(x) = $\binom{5}{2}*2^{n-5}$ = $\frac{5}{16}*2^{n}$

$\frac{5}{16} < \frac{3}{8}$

after that value of f(x) won't increase. So max value of f(x) = $\frac{3}{8}*2^{n}$

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