Question

设a为正实数,函数f(x)=x*3-ax*2-a*2x+1,x属于全体实数,求f(x)的极值

Answer 1

f(x) = x^3-ax^2-a^2x+1

f'(x) = 3x^2-2ax-a^2 = 0

(3x+a)(x-a) = 0

x = a or -a/3

f''(x) =6x-2a

f''(a) = 6a -2a = 4a > 0 ( min)

f''(-a/3) = -2a-2a = -4a < 0 ( max)

f(a) = -a^3 + 1

f(-a/3) = -a^3/27 - a^3/9 + a^3/3 +1

= 5a^3/27 +1

maxf(x) = f(-a/3) = 5a^3/27 +1

minf(x) = f(a) = -a^3 + 1