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受欢迎的函数 >

critical sin(2x),0<= x<= 2pi

解答

临界点

sin (2 x), 0 \leq x \leq 2 π

解答

x = 0, x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}, x = 2 π

求解步骤

sin (2 x), 0 \leq x \leq 2 π

找到

f^{'} (x)

等于零或无定义的点

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

确定不在

f (x)

定义域内的临界点

sin (2 x), 0 \leq x \leq 2 π

的定义域

: 0 \leq x \leq 2 π

所有临界点均在定义域内

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

计算区间端点

x = 0 :

临界点

计算区间端点

x = 2 π :

临界点

x = 0, x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}, x = 2 π

作图

流行的例子

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cr i t i c a l f (x) = - x^{2} + 4

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