extreme f(x)=9t^2+4t-10

解答

极值点

f (x) = 9 t^{2} + 4 t - 10

极小值 (- \frac{2}{9}, - \frac{94}{9})

求解步骤

f^{'} (t) = 18 t + 4

Find intervals:

递减

: - \infty < t < - \frac{2}{9},

递增

: - \frac{2}{9} < t < \infty

将

t = - \frac{2}{9}

代入

9 t^{2} + 4 t - 10 : - \frac{94}{9}

极小值 (- \frac{2}{9}, - \frac{94}{9})

e x t re m e (x^{2} - 4)^{\frac{1}{7}}, - 2 \leq x \leq 3

e x t re m e f (x) = \frac{3}{2} t^{4} - t^{6}

e x t re m e f (x, y) = - x^{2} - y^{3} + 3 y^{2} - 2 x + 45 y + 1

e x t re m e y = - 12 x^{5} - 45 x^{4} + 20 x^{3} - 1

e x t re m e f (x) = x^{3} - 2 x - 4 x + 4 (0.3)