(b) $\frac{e^{x^2} \sqrt{\sin x}}{(2 x+1)^3}$
Answer (2)
The problem asks to find the domain of the function ( 2 x + 1 ) 3 e x 2 s i n x .
The square root requires sin x ≥ 0 , which occurs when 2 nπ ≤ x ≤ ( 2 n + 1 ) π for any integer n .
The denominator requires x = − 2 1 .
The domain is the union of the intervals [ 2 nπ , ( 2 n + 1 ) π ] for all integers n .
Explanation
Problem Analysis We are given the expression ( 2 x + 1 ) 3 e x 2 s i n x and asked to analyze it, which primarily involves determining its domain. The domain of a function is the set of all possible input values (x-values) for which the function is defined.
Identifying Restrictions To find the domain, we need to consider the restrictions imposed by each part of the expression:
sin x : The square root function is only defined for non-negative values. Therefore, we must have sin x ≥ 0 .
( 2 x + 1 ) 3 in the denominator: The denominator cannot be zero, so 2 x + 1 = 0 , which means x = − 2 1 .
e x 2 : The exponential function is defined for all real numbers, so it doesn't impose any restrictions on the domain.
Analyzing sin(x) >= 0 Now, let's analyze the condition sin x ≥ 0 . The sine function is non-negative in the intervals [ 2 nπ , ( 2 n + 1 ) π ] , where n is an integer. This means that x must lie in one of these intervals.
For example:
When n = 0 , we have [ 0 , π ] .
When n = 1 , we have [ 2 π , 3 π ] .
When n = − 1 , we have [ − 2 π , − π ] .
Checking x = -1/2 Next, we need to check if x = − 2 1 falls within any of the intervals where sin x ≥ 0 . Since − 2 π ≈ − 6.28 and − π ≈ − 3.14 , the interval [ − 2 π , − π ] is approximately [ − 6.28 , − 3.14 ] . Since − 0.5 is not within this range, x = − 2 1 does not fall within the interval [ − 2 π , − π ] .
However, we need to consider the interval containing 0, which is [ 0 , π ] . Since − 2 1 is not in this interval either, we don't need to exclude it from the intervals where sin x ≥ 0 .
Determining the Domain Therefore, the domain of the given expression is the union of the intervals [ 2 nπ , ( 2 n + 1 ) π ] for all integers n , excluding the point x = − 2 1 if it happens to fall within any of these intervals. In this case, since − 2 1 does not fall within the intervals where sin x ≥ 0 around 0, we don't need to explicitly exclude it.
Final Answer In conclusion, the domain of the function ( 2 x + 1 ) 3 e x 2 s i n x is the set of all x such that 2 nπ ≤ x ≤ ( 2 n + 1 ) π for any integer n .
Examples
Understanding the domain of a function is crucial in many real-world applications. For example, when modeling the height of a projectile over time using a function, the domain would represent the valid time intervals for which the model makes sense (e.g., time cannot be negative). Similarly, in electrical engineering, when analyzing the behavior of an alternating current circuit, the domain of the voltage or current function would represent the time intervals over which the circuit is operating. In both cases, knowing the domain ensures that the function's output is meaningful and physically realistic.
The domain of the function ( 2 x + 1 ) 3 e x 2 s i n x is all x in the intervals [ 2 nπ , ( 2 n + 1 ) π ] for any integer n , excluding x = − 2 1 . The sine function must be non-negative, and the denominator must be non-zero. Thus, the domain comprehensively covers the permissible intervals where these conditions are met.
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