Question

I have a cubic function that is x^3+ax^2+bx+c is increasing on (negative infinity , positive infinity) if b > a^2/3. would I just take the derivative to show this?

Answer 1

Sure, you'll get $$f'(x)=3x^2+2ax+b$$ and you want $$f'>0$$ so:$$3x^2+2ax+b>0$$If we have zeros they will be at either $$x=\frac{-2a\pm\sqrt{4a^2-12b}}6=\frac{-a\pm\sqrt{a^2-3b}}{3}$$Observe that \(b>a^2/3\) means \(a^2-3b<0\) hence we have no real zeros and \(f'>0\) (we know to eliminate the possibility \(f'<0\) because \(f'''=6>0\) and our derivative has positive curvature)