Prove that for all x\geq 1\Longrightarrow x^4-4x+3\geq 0 Any ideas how to start?
K kanali New member Joined Jun 12, 2010 Messages 18 Jun 25, 2010 #1 Prove that for all x≥1⟹x4−4x+3≥0\displaystyle x\geq 1\Longrightarrow x^4-4x+3\geq 0x≥1⟹x4−4x+3≥0 Any ideas how to start?
Prove that for all x≥1⟹x4−4x+3≥0\displaystyle x\geq 1\Longrightarrow x^4-4x+3\geq 0x≥1⟹x4−4x+3≥0 Any ideas how to start?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Jun 25, 2010 #2 Did you try factoring the given quartic?. (x−1)2(x2+2x+3)\displaystyle (x-1)^{2}(x^{2}+2x+3)(x−1)2(x2+2x+3)
Did you try factoring the given quartic?. (x−1)2(x2+2x+3)\displaystyle (x-1)^{2}(x^{2}+2x+3)(x−1)2(x2+2x+3)
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Jun 25, 2010 #3 One way, to wit: x4−4x+3 = (x−1)2(x2+2x+3) and x−1 ≥ 0.\displaystyle One \ way, \ to \ wit: \ x^4-4x+3 \ = \ (x-1)^2(x^2+2x+3) \ and \ x-1 \ \ge \ 0.One way, to wit: x4−4x+3 = (x−1)2(x2+2x+3) and x−1 ≥ 0.
One way, to wit: x4−4x+3 = (x−1)2(x2+2x+3) and x−1 ≥ 0.\displaystyle One \ way, \ to \ wit: \ x^4-4x+3 \ = \ (x-1)^2(x^2+2x+3) \ and \ x-1 \ \ge \ 0.One way, to wit: x4−4x+3 = (x−1)2(x2+2x+3) and x−1 ≥ 0.
K kanali New member Joined Jun 12, 2010 Messages 18 Jun 25, 2010 #4 galactus said: Did you try factoring the given quartic?. (x−1)2(x2+2x+3)\displaystyle (x-1)^{2}(x^{2}+2x+3)(x−1)2(x2+2x+3) Click to expand... Thanks i get it now
galactus said: Did you try factoring the given quartic?. (x−1)2(x2+2x+3)\displaystyle (x-1)^{2}(x^{2}+2x+3)(x−1)2(x2+2x+3) Click to expand... Thanks i get it now
