The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
0 votes
Find the maximum value of f(x). n is a constant

f(x)= C(x,2) *$2^{n-x}$
asked in Calculus by Boss (24.4k points) | 178 views
$\frac{3}{8} * 2^{n}$ ?
yes. can you share you solution. ?I am doing some mistake and not able to find:(
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) = $\frac{3}{8}*2^{n}$

Please log in or register to answer this question.

Related questions

0 votes
1 answer
asked Nov 28, 2018 in Calculus by aditi19 Active (3.7k points) | 148 views
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
49,781 questions
54,511 answers
75,107 users