方程x^3-6x^2+9x-10=0的实根个数是多少

Question

如题所述

Answer 1

f'(x)=3x²-12x+9=3(x²-4x+3=3(x-3)(x-1)
令f'(x)≤0
3(x-3)(x-1)≤0
1≤x≤3
f(x)在区间[1,3]上单调递减,在(-∞,1]上和[3,+∞)上都单调递增.
f(1)=1-6+9-10=-6<0,f(x)在(-∞,1]上恒<0
f(x)在[1,3]上单调递减,在[1,3]上恒<0
f(x)在[3,+∞)上单调递增,f(10)=1000-600+90-10=480>0
又f(x)连续,f(x)在区间[3,+∞)上和x轴有且仅有一个交点.
方程的实数根的个数是1,即仅有一个实根.