Question

已知函数f x=-x的三次幂+ax的二次幂-4在x=2处取得极值，若m,n属于[-1,1]，则

Answer 1

f(x)=-x^3+ax^2-4f'(x)=-3x^2+2axf'(2)=-12+4a=0a=3f(x)=-x^3+3x^2-4f'(x)=-3x^2+6xf(x)在x=2和x=0处有极点,当-1≤x≤0时,f'(x)≤0,f(x)单调递减,f(0)≤f(x)≤f(-1)-4≤f(x)≤0;当0≤x≤1时,f'(x)≥0,f'(x)单调递增,f(0)≤f(x)≤f(1);-4≤f(x)≤-2;所以当m=0时,f(m)取最小值-4;f''(x)=-6x+6f'(x)在x=1处有极点,当-1≤x≤1时,f''(x)≥0,f'(x)单调递增,-9≤f'(x)≤3;所以当n=-1时f'(n)取最小值为-9.所以f(m)+f'(n)的最小值为-4-9=-13