Hi!

Or if we evaluate f(-1) by the remainder theorem the signs in the quotient followed by the
remainder strictly alternate in sign:  + - + - + - + so there are no negative roots less than or
equal to -1.  (Lower bound theorem)

Similarly if we evaluate f(1) by the remainder theorem the signs in the quotient followed by
the remainder are all positive.  So there are no positive roots greater than or equal to 1.
(Upper bound theorem)

If |x|<=1 we have  x+3>=2.  Also the even powered terms cannot be negative.   Hence the
complete sum is >= 2 for all x in [-1,1]. 

Hence f(x)=0 has no real roots, and in fact is always positive since if there were a value a outside
of [-1,1] for which f(a)<0 then the intermediate value theorem would guarantee a root between
either 1 and a if a>1 or between -1 and a if a<-1.


Or more elegantly along the lines that Jack pointed out:  Completing the square on x^2+x+3 we get

x^2+x+3 = (x+1/2)^2 - 1/4 + 3 = (x+1/2)^2 + 11/4 which has minimum 11/4.

Also x^6 + x^4 is always >= zero.  Hence f(x) >= 11/4 (actually strictly >) for all x in the reals.
f(-1/2) = 181/65 = 2.828125.

Have a very blessed day!

Thanks bobym and bob bundy

I have solved it like this way....

Sorry, (-7,7), not closed interval.

hi jacks,

I thought I'd start by looking at the graph y = F(x)

It has a single minimum as you can see and y > 0 for all x.

That's enough to show it has no real roots.

But a graph alone doesn't constitute a proof as it relies on a tool and doesn't show all values of x.

So one approach would be to prove that the graph truely has those properties.

Bob

Prove that the equation