Question

Solveeeeeeeeeeeeeee

if p(x) = px^7 + qx^3 + rx - 5
where p , q , r are constants , if p(-7) = 7
then p(7) = ?

Answer 1

not enough info to give a real number for f(7)

Answer 2

however we can give the answer in terms of p,q,r

Answer 3

how? and it's p(7) not f(7)

Answer 4

this looks like an odd function

Answer 5

yeah p(7)

Answer 6

lets start with plugging in p\(-7)=7

Answer 7

I don't know if this works..
p(x) = px^7 + qx^3 + rx - 5
p(-7) = p(-7)^7 + q(-7)^3 + r(-7) - 5 = 7
p(-7)^7 + q(-7)^3 + r(-7) = 12
- [ p(7)^7 + (q(7)^3 + r(7)] = 12
p(7)^7 + (q(7)^3 + r(7) = -12
So,
p(7) = p(7)^7 + q(7)^3 + r(7) - 5
Plug in the results for p(7)^7 + (q(7)^3 + r(7) , I think we can get the answer..

Answer 8

\[-7^7p-7^3q-7r-5=7 \implies -7^7p-7^3q-7r=12\]
\[p(7)=7^7p+7^3q+7+5=-12+5=7\]

Answer 9

@RolyPoly
Yes, i m done same procedure, but i at end i m stuck

Answer 10

some how we got the answer because of symmetry

Answer 11

@msingh
Stuck? where?

Answer 12

wait u have done some where wrong @RolyPoly

Answer 13

p(7)=7

Answer 14

p(x) = px^7 + qx^3 + rx - 5
p(-7) = p(-7)^7 + q(-7)^3 + r(-7) - 5 = 7
p(-7)^7 + q(-7)^3 + r(-7) = 12

- [ p(7)^7 + (q(7)^3 + r(7)] = 12

p(7)^7 + (q(7)^3 + r(7) = -12

How can u get minus sign common, it's impossible

Answer 15

@Jonask
theres one mistake in your calculations

Answer 16

p(7)=7^7p+7^3q+7-5=-12-5=-17

Answer 17

the function is neither even nor odd ,

Answer 18

oh yeah i made a mistake

Answer 19

@Jonask I don't think giving out answer makes you cool

p(-7)^7 + q(-7)^3 + r(-7) = 12
For the first term: p (-7)^3 = p(-7)(-7)(-7)...(-7) = -p(7)^7
For the second term:q(-7)^3 = q (-7)(-7)(-7) = -q(7)^3
So, take out -1 as the common factor.
- [ p(7)^7 + (q(7)^3 + r(7)] = 12

Answer 20

no worries, i wont take away your medal :)

Answer 21

p(x) = px^7 + qx^3 + rx - 5
p(-7) = p(-7)^7 + q(-7)^3 + r(-7) - 5 = 7
p(-7)^7 + q(-7)^3 -7r = 12

Answer 22

@Jonask Giving out answer doesn't make you cool

Answer 23

@RolyPoly
thats rude rolypoly, he gave an explanation

Answer 24

thanks for the knowledge ,i know also its part of the openstudy code of conduct 4 give me

Answer 25

whats openstudy code of conduct 4,

Answer 26

http://openstudy.com/code-of-conduct

Answer 27

I believe Jonask guided the person

Answer 28

i have only one medal otherwise i wud have given to all of u one one, its not something to fight on, everyone know the answer some answer it some not.

Answer 29

I'm not here for medals.

Answer 30

do you guys agree
the answer is
p(7)= -17

Answer 31

yes

Answer 32

by the way, even if p(7) = 7, that is not enough to make the function symmetric. because it could be a coincidence, you need all elements in your domain p(-x) = p(x) for even
p(-x) = - p(x) for odd.
now do you see why its neither even nor odd. its a sum of an odd function plus an even function

Answer 33

@perl
how it came -17

Answer 34

one sec

Answer 35

p(x) = px^7 + qx^3 + rx - 5
p(-7) = p(-7)^7 + q(-7)^3 + r(-7) - 5 = 7
p(-7)^7 + q(-7)^3 + r(-7) = 12

- [ p(7)^7 + (q(7)^3 + r(7)] = 12
so we know that
p(7)^7 + (q(7)^3 + r(7) = -12
now find p(7)

p(7) = p(7)^7 + (q(7)^3 + r(7) -5 , and substitute from above
= ( -12) - 5

Answer 36

thank u