0568 #17

gaucho85 · Post by **gaucho85** » Thu Oct 16, 2008 2:26 pm

how many real roots does 2x^5 +8x -7 have?

Anyone have any fancy ways to solove this?

I looked for stationary points by setting the derrivative to 0, to get 10x^4+8=0 so x^4=-4/5

Since x^4>0 there are no real stationary points so there must only be one real root?

It seems like it works on this problem, but only because there were no stationary points? Could they ask a problem with stationary points? Would you just have to investigate between those points?

Thanks

moo5003 · Post by **moo5003** » Thu Oct 16, 2008 3:47 pm

A question like this could proabably show up (I see no reason why it couldnt) though be careful, just because the derivative is always positive does not imply there is one root. Its possible for there to be 0 roots as well. You must check it has a negative value first.

For Example: If the function only approached the x axis instead of going below it.

Nameless · Post by **Nameless** » Thu Oct 16, 2008 4:09 pm

how many real roots does 2x^5 +8x -7 have?

Anyone have any fancy ways to solove this?

I looked for stationary points by setting the derrivative to 0, to get 10x^4+8=0 so x^4=-4/5

Since x^4>0 there are no real stationary points so there must only be one real root?

It seems like it works on this problem, but only because there were no stationary points? Could they ask a problem with stationary points? Would you just have to investigate between those points?

Thanks

Since, the degree is 5 - ODD
so f(x)------->+infinity when x----->+infinity
and f(x)------->-infinity when x----->-infinity
so we can find M, N st f(M)f(N)<0 so f(x)=0 has at least one root
Moreover, f'(x)=4x^2+8 >0 then f(x) has only one root