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 \(\displaystyle x\geq 1\Longrightarrow x^4-4x+3\geq 0\) 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?. \(\displaystyle (x-1)^{2}(x^{2}+2x+3)\)
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Jun 25, 2010 #3 \(\displaystyle One \ way, \ to \ wit: \ x^4-4x+3 \ = \ (x-1)^2(x^2+2x+3) \ and \ x-1 \ \ge \ 0.\)
K kanali New member Joined Jun 12, 2010 Messages 18 Jun 25, 2010 #4 galactus said: Did you try factoring the given quartic?. \(\displaystyle (x-1)^{2}(x^{2}+2x+3)\) Click to expand... Thanks i get it now
galactus said: Did you try factoring the given quartic?. \(\displaystyle (x-1)^{2}(x^{2}+2x+3)\) Click to expand... Thanks i get it now