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受欢迎的三角函数 >

2arcsin(x^2-2x+(sqrt(3))/2)>(3pi)/2

解答

2 arcsin (x^{2} - 2 x + \frac{3}{2}) > \frac{3 π}{2}

解答

对所有 x \in R 为假

求解步骤

2 arcsin (x^{2} - 2 x + \frac{3}{2}) > \frac{3 π}{2}

两边除以

2

2 arcsin (x^{2} - 2 x + \frac{3}{2}) > \frac{3 π}{2}

两边除以

2

\frac{2 arcsin ( x ^{2} - 2 x + \frac{3}{2} )}{2} > \frac{\frac{3 π}{2}}{2}

化简

\frac{2 arcsin ( x ^{2} - 2 x + \frac{3}{2} )}{2} > \frac{\frac{3 π}{2}}{2}

化简

\frac{2 arcsin ( x ^{2} - 2 x + \frac{3}{2} )}{2} : arcsin (x^{2} - 2 x + \frac{3}{2})

\frac{2 arcsin ( x ^{2} - 2 x + \frac{3}{2} )}{2}

数字相除:

\frac{2}{2} = 1

= arcsin (x^{2} - 2 x + \frac{3}{2})

化简

\frac{\frac{3 π}{2}}{2} : \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

arcsin (x^{2} - 2 x + \frac{3}{2}) > \frac{3 π}{4}

arcsin (x^{2} - 2 x + \frac{3}{2}) > \frac{3 π}{4}

arcsin (x^{2} - 2 x + \frac{3}{2}) > \frac{3 π}{4}

arcsin (x^{2} - 2 x + \frac{3}{2})

的值域:

arcsin (\frac{3}{2} - 1) \leq arcsin (x^{2} - 2 x + \frac{3}{2}) \leq \frac{π}{2}

函数值域定义

x^{2} - 2 x + \frac{3}{2}

的值域:

f (x) \geq \frac{3}{2} - 1

函数值域定义

确定每个定义区间的极小值和极大值并将结果合并

x^{2} - 2 x + \frac{3}{2}

的定义域

:

对所有

x \in R

为真

定义域定义

函数没有无定义点也没有定义域限制。因此，定义域为

对所有 x \in R 为真

x^{2} - 2 x + \frac{3}{2}

的极值点:

极小值

(1, \frac{3}{2} - 1)

一阶导数判别法的定义

f^{'} (x) = 2 x - 2

\frac{d}{d x} (x^{2} - 2 x + \frac{3}{2})

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (x^{2}) - \frac{d}{d x} (2 x) + \frac{d}{d x} (\frac{3}{2})

\frac{d}{d x} (x^{2}) = 2 x

\frac{d}{d x} (x^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 x^{2 - 1}

化简

= 2 x

\frac{d}{d x} (2 x) = 2

\frac{d}{d x} (2 x)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d x}{d x}

使用常见微分定则:

\frac{d x}{d x} = 1

= 2 \cdot 1

化简

= 2

\frac{d}{d x} (\frac{3}{2}) = 0

\frac{d}{d x} (\frac{3}{2})

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 2 x - 2 + 0

化简

= 2 x - 2

Find intervals:

递减

: - \infty < x < 1,

递增

: 1 < x < \infty

f^{'} (x) = 2 x - 2

找到临界点:

x = 1

临界点定义

f^{'} (x) = 0 : x = 1

2 x - 2 = 0

将

2

到右边

2 x - 2 = 0

两边加上

2

2 x - 2 + 2 = 0 + 2

化简

2 x = 2

2 x = 2

两边除以

2

2 x = 2

两边除以

2

\frac{2 x}{2} = \frac{2}{2}

化简

x = 1

x = 1

x = 1

f^{'} (x) > 0 : x > 1

2 x - 2 > 0

将

2

到右边

2 x - 2 > 0

两边加上

2

2 x - 2 + 2 > 0 + 2

化简

2 x > 2

2 x > 2

两边除以

2

2 x > 2

两边除以

2

\frac{2 x}{2} > \frac{2}{2}

化简

x > 1

x > 1

f^{'} (x) < 0 : x < 1

2 x - 2 < 0

将

2

到右边

2 x - 2 < 0

两边加上

2

2 x - 2 + 2 < 0 + 2

化简

2 x < 2

2 x < 2

两边除以

2

2 x < 2

两边除以

2

\frac{2 x}{2} < \frac{2}{2}

化简

x < 1

x < 1

将区间与定义域合并

x^{2} - 2 x + \frac{3}{2}

的定义域

:

对所有

x \in R

为真

定义域定义

函数没有无定义点也没有定义域限制。因此，定义域为

对所有 x \in R 为真

将

x = 1

与定义域合并:

x = 1

x = 1 and 对所有 x \in R 为真

化简

x = 1

将

1 < x < \infty

与定义域合并:

x > 1

1 < x < \infty and 对所有 x \in R 为真

化简

x > 1

将

- \infty < x < 1

与定义域合并:

x < 1

- \infty < x < 1 and 对所有 x \in R 为真

化简

x < 1

- \infty < x < 1, x = 1, 1 < x < \infty

- \infty < x < 1, x = 1, 1 < x < \infty

单调区间行为特性汇总

符号 行为 - \infty < x < 1 f^{'} (x) < 0 递减 x = 1 f^{'} (x) = 0 极小值 1 < x < \infty f^{'} (x) > 0 递增

递减 : - \infty < x < 1, 递增 : 1 < x < \infty

将

x = 1

代入

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

1^{2} - 2 \cdot 1 + \frac{3}{2}

化简

\frac{3}{2} - 1

极小值 (1, \frac{3}{2} - 1)

确定

- \infty < x < \infty

区间对应的值域:

\frac{3}{2} - 1 \leq f (x) < \infty

计算函数在区间边缘处的值:

x \to - \infty lim (x^{2} - 2 x + \frac{3}{2}) = \infty

x \to - \infty lim (x^{2} - 2 x + \frac{3}{2})

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

不定型除外

= x \to - \infty lim (x^{2}) - x \to - \infty lim (2 x) + x \to - \infty lim (\frac{3}{2})

x \to - \infty lim (x^{2}) = \infty

x \to - \infty lim (x^{2})

运用无穷大属性:

x \to \pm \infty lim (a x^{n} + \dots + b x + c) = \infty, a > 0,

n is even

a = 1, n = 2

= \infty

x \to - \infty lim (2 x) = - \infty

x \to - \infty lim (2 x)

运用无穷大属性:

x \to - \infty lim (a x^{n} + \dots + b x + c) = - \infty, a > 0,

n is odd

a = 2, n = 1

= - \infty

x \to - \infty lim (\frac{3}{2}) = \frac{3}{2}

x \to - \infty lim (\frac{3}{2})

x \to a lim c = c

= \frac{3}{2}

= \infty - (- \infty) + \frac{3}{2}

化简

\infty - (- \infty) + \frac{3}{2} : \infty

\infty - (- \infty) + \frac{3}{2}

运用无穷大属性:

\infty + \infty = \infty

= \infty + \frac{3}{2}

运用无穷大属性:

\infty + c = \infty

= \infty

= \infty

x \to \infty lim (x^{2} - 2 x + \frac{3}{2}) = \infty

x \to \infty lim (x^{2} - 2 x + \frac{3}{2})

使用以下的代数特性

: a + b = a (1 + \frac{b}{a})

x^{2} - 2 x + \frac{3}{2} = x^{2} (1 - \frac{2}{x} + \frac{3}{2 x ^{2}})

= x \to \infty lim (x^{2} (1 - \frac{2}{x} + \frac{3}{2 x ^{2}}))

x \to a lim [f (x) \cdot g (x)] = x \to a lim f (x) \cdot x \to a lim g (x)

不定型除外

= x \to \infty lim (x^{2}) \cdot x \to \infty lim (1 - \frac{2}{x} + \frac{3}{2 x ^{2}})

x \to \infty lim (x^{2}) = \infty

x \to \infty lim (x^{2})

运用无穷大属性:

x \to \pm \infty lim (a x^{n} + \dots + b x + c) = \infty, a > 0,

n is even

a = 1, n = 2

= \infty

x \to \infty lim (1 - \frac{2}{x} + \frac{3}{2 x ^{2}}) = 1

x \to \infty lim (1 - \frac{2}{x} + \frac{3}{2 x ^{2}})

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

不定型除外

= x \to \infty lim (1) - x \to \infty lim (\frac{2}{x}) + x \to \infty lim (\frac{3}{2 x ^{2}})

x \to \infty lim (1) = 1

x \to \infty lim (1)

x \to a lim c = c

= 1

x \to \infty lim (\frac{2}{x}) = 0

x \to \infty lim (\frac{2}{x})

运用无穷大属性:

x \to \infty lim (\frac{c}{x ^{a}}) = 0

= 0

x \to \infty lim (\frac{3}{2 x ^{2}}) = 0

x \to \infty lim (\frac{3}{2 x ^{2}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{3}{2} \cdot x \to \infty lim (\frac{1}{x ^{2}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

不定型除外

= \frac{3}{2} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( x ^{2} )}

x \to \infty lim (1) = 1

x \to \infty lim (1)

x \to a lim c = c

= 1

x \to \infty lim (x^{2}) = \infty

x \to \infty lim (x^{2})

运用无穷大属性:

x \to \pm \infty lim (a x^{n} + \dots + b x + c) = \infty, a > 0,

n is even

a = 1, n = 2

= \infty

= \frac{3}{2} \cdot \frac{1}{\infty}

化简

\frac{3}{2} \cdot \frac{1}{\infty} : 0

\frac{3}{2} \cdot \frac{1}{\infty}

运用无穷大属性:

\frac{c}{\infty} = 0

= \frac{3}{2} \cdot 0

使用法则

0 \cdot a = 0

= 0

= 0

= 1 - 0 + 0

化简

= 1

= \infty \cdot 1

运用无穷大属性:

c \cdot \infty = \infty

= \infty

该区间在

x = 1

处有极小值点，极小值为

f (1) = \frac{3}{2} - 1

将边缘处的函数值与函数在区间内的极值点合并:

在定义域区间

- \infty < x < \infty

处的最小函数值为

\frac{3}{2} - 1

在定义域区间

- \infty < x < \infty

处的最大函数值为

\infty

因此

x^{2} - 2 x + \frac{3}{2}

在定义域区间

- \infty < x < \infty

上的值域为

\frac{3}{2} - 1 \leq f (x) < \infty

将所有定义域区间对应的值域合并得出函数值域

f (x) \geq \frac{3}{2} - 1

因为

arcsin

为递增函数，值域为

- \frac{π}{2} \leq arcsin (x) \leq \frac{π}{2}

和

x^{2} - 2 x + \frac{3}{2} \geq \frac{3}{2} - 1

arcsin (\frac{3}{2} - 1) \leq arcsin (x^{2} - 2 x + \frac{3}{2}) \leq \frac{π}{2}

arcsin (x^{2} - 2 x + \frac{3}{2}) > \frac{3 π}{4} and arcsin (\frac{3}{2} - 1) \leq arcsin (x^{2} - 2 x + \frac{3}{2}) \leq \frac{π}{2} :

假

令

y = arcsin (x^{2} - 2 x + \frac{3}{2})

合并区间

y > \frac{3 π}{4} and arcsin (\frac{3}{2} - 1) \leq y \leq \frac{π}{2}

合并重叠的区间

y > \frac{3 π}{4} and arcsin (\frac{3}{2} - 1) \leq y \leq \frac{π}{2}

两个区间的交集是指同时存在于这两个区间的数的集合

y > \frac{3 π}{4}

and

arcsin (\frac{3}{2} - 1) \leq y \leq \frac{π}{2}

对所有 y \in R 为假

对所有 y \in R 为假

x \in R 无解

对所有 x \in R 为假

流行的例子

pi/2-arctan(e^x)<0.01

\frac{π}{2} - arctan (e^{x}) < 0.01

2>(24)/(sin(θ))

2 > \frac{24}{sin ( θ )}

5sin(1/2 (x+pi/4))-1>= 7

5 sin (\frac{1}{2} (x + \frac{π}{4})) - 1 \geq 7

arctan(x)<= 10^3

arctan (x) \leq 1 0^{3}

4 sin^{2} (x) \geq 1