polynomials (1 Viewer)

[ninetypercent]

my turn :]

Prove that P(x) = , has no real zeros if c > 28/3

Thanks

 

[Affinity]

Differentiate to get P'(x)

solve P'(x) = 0

then plug back into P(x),
etc.

 

[Lukybear]

Differentiate to get P'(x)

solve P'(x) = 0

then plug back into P(x),
etc.

Could you expand on that?

 

[Affinity]

P'(x) = x^3 - x^2 - 4x + 4
= x^2(x - 1) - 4(x - 1)
= (x^2 - 4)(x - 1)
= (x+2)(x-2)(x-1)

so P'(x) = 0 for x =2,1,-2

P(1) = 23/12 + c
P(2) = 4/3 + c
P(-2) = c - 28/3

so P(x) > 0 for all stationary points (hence all x since P increases without bounds for large and small x) and
has no real roots (otherwise a real root exist because you will have P(a), > 0 and P(b) < 0 for some a,b
precisely when all of the above are > 0 ic C > 28/3

 

[Lukybear]

P'(x) = x^3 - x^2 - 4x + 4
= x^2(x - 1) - 4(x - 1)
= (x^2 - 4)(x - 1)
= (x+2)(x-2)(x-1)

so P'(x) = 0 for x =2,1,-2

P(1) = 23/12 + c
P(2) = 4/3 + c
P(-2) = c - 28/3

so P(x) > 0 for all stationary points (hence all x since P increases without bounds for large and small x) and
has no real roots (otherwise a real root exist because you will have P(a), > 0 and P(b) < 0 for some a,b
precisely when all of the above are > 0 ic C > 28/3

Thanks very much!