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Popular >

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

Solution

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

Solution

False for all x \in R

Solution steps

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

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

Simplify

\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}

Divide the numbers:

\frac{2}{2} = 1

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

Simplify

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

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

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}

Range of

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

Function range definition

Range of

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

Function range definition

Find the minimum and maximum value in each defined interval and unite the results

Domain of

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

True for all

x \in R

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

True for all x \in R

Extreme Points of

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

Minimum

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

First Derivative Test definition

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

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

Apply the Sum/Difference Rule:

(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})

Apply the Power Rule:

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

= 2 x^{2 - 1}

Simplify

= 2 x

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

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

Take the constant out:

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

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

Apply the common derivative:

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

= 2 \cdot 1

Simplify

= 2

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

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

Derivative of a constant:

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

= 0

= 2 x - 2 + 0

Simplify

= 2 x - 2

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < \infty

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

Find the critical points:

x = 1

Critical point definition

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

2 x - 2 = 0

Move

2

to the right side

2 x - 2 = 0

Add

2

to both sides

2 x - 2 + 2 = 0 + 2

Simplify

2 x = 2

2 x = 2

Divide both sides by

2

2 x = 2

Divide both sides by

2

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

Simplify

x = 1

x = 1

x = 1

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

2 x - 2 > 0

Move

2

to the right side

2 x - 2 > 0

Add

2

to both sides

2 x - 2 + 2 > 0 + 2

Simplify

2 x > 2

2 x > 2

Divide both sides by

2

2 x > 2

Divide both sides by

2

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

Simplify

x > 1

x > 1

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

2 x - 2 < 0

Move

2

to the right side

2 x - 2 < 0

Add

2

to both sides

2 x - 2 + 2 < 0 + 2

Simplify

2 x < 2

2 x < 2

Divide both sides by

2

2 x < 2

Divide both sides by

2

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

Simplify

x < 1

x < 1

Combine intervals with domain

Domain of

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

True for all

x \in R

Domain definition

The function has no undefined points nor domain constraints. Therefore, the domain is

True for all x \in R

Combine

x = 1

with domain:

x = 1

x = 1 and True for all x \in R

Simplify

x = 1

Combine

1 < x < \infty

with domain:

x > 1

1 < x < \infty and True for all x \in R

Simplify

x > 1

Combine

- \infty < x < 1

with domain:

x < 1

- \infty < x < 1 and True for all x \in R

Simplify

x < 1

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

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

Summary of the monotone intervals behavior

Sign Behavior - \infty < x < 1 f^{'} (x) < 0 Decreasing x = 1 f^{'} (x) = 0 Minimum 1 < x < \infty f^{'} (x) > 0 Increasing

Decreasing : - \infty < x < 1, Increasing : 1 < x < \infty

Plug

x = 1

into

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

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

Simplify

\frac{3}{2} - 1

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

Find the range for the interval

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

Compute the values of the function at the edges of the interval:

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)

With the exception of indeterminate form

= 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})

Apply Infinity Property:

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)

Apply Infinity Property:

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}

Simplify

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

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

Apply Infinity Property:

\infty + \infty = \infty

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

Apply Infinity Property:

\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})

Apply the following algebraic property

: 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)

With the exception of indeterminate form

= 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})

Apply Infinity Property:

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)

With the exception of indeterminate form

= 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})

Apply Infinity Property:

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

With the exception of indeterminate form

= \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})

Apply Infinity Property:

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}

Simplify

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

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

Apply Infinity Property:

\frac{c}{\infty} = 0

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

Apply rule

0 \cdot a = 0

= 0

= 0

= 1 - 0 + 0

Simplify

= 1

= \infty \cdot 1

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

The interval has a minimum point at

x = 1

with value

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

Combine the function value at the edge with the extreme points of the function in the interval:

Minimum function value at the domain interval

- \infty < x < \infty

\frac{3}{2} - 1

Maximum function value at the domain interval

- \infty < x < \infty

\infty

Therefore the range of

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

at the domain interval

- \infty < x < \infty

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

Combine the ranges of all domain intervals to obtain the function range

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

Since

arcsin

is an increasing function with range of

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

and

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} :

False

Let

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

Combine the intervals

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

Merge Overlapping Intervals

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

The intersection of two intervals is the set of numbers which are in both intervals

y > \frac{3 π}{4}

and

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

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

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