令 f(x) = x^2-2x+m-1 ,此抛物线开口向上.
要使一个根在(-2,0)内,一个在(1,3)内,则有:
① f(-2) = m+7 > 0 ,解得:m > -7 ;
② f(0) = m-1 < 0 ,解得:m < 1 ;
③ f(1) = m-2 < 0 ,解得:m < 2 ;
④ f(3) = m+2 > 0 ,解得:m > -2 ;
综上可得:-2 < m < 1 .
令 f(x) = x^2-2x+m-1 ,此抛物线开口向上.
要使一个根在(-2,0)内,一个在(1,3)内,则有:
① f(-2) = m+7 > 0 ,解得:m > -7 ;
② f(0) = m-1 < 0 ,解得:m < 1 ;
③ f(1) = m-2 < 0 ,解得:m < 2 ;
④ f(3) = m+2 > 0 ,解得:m > -2 ;
综上可得:-2 < m < 1 .