extreme f(x)=x^2+3x+1

解答

极值点

f (x) = x^{2} + 3 x + 1

极小值 (- \frac{3}{2}, - \frac{5}{4})

求解步骤

f^{'} (x) = 2 x + 3

Find intervals:

递减

: - \infty < x < - \frac{3}{2},

递增

: - \frac{3}{2} < x < \infty

将

x = - \frac{3}{2}

代入

x^{2} + 3 x + 1 : - \frac{5}{4}

极小值 (- \frac{3}{2}, - \frac{5}{4})

e x t re m e x - 3 x

e x t re m e f (x) = 3 x + 1 + \frac{1}{x - 1}

e x t re m e f (x) = 9 x^{5} - 2 x^{4} - 2 x^{3} + 4

e x t re m e 3 x^{2} - 12 x - 15

e x t re m e f (x) = x y + \frac{9}{x} + \frac{3}{y}