Functions Domains And Equation Solving
Functions are the fundamental building blocks of mathematics, mapping inputs to outputs according to specific rules. Understanding the behavior of different types of functions is crucial in various fields, from calculus and analysis to computer science and engineering. In this article, we will delve into the properties of several functions, including logarithmic, square root, polynomial, exponential, and reciprocal functions. We will explore their domains, which define the set of permissible input values, and how to solve equations involving these functions. Understanding the domain of a function is paramount as it dictates the values for which the function is valid and produces meaningful outputs. Let's consider the following functions:
- f(x) = (1/3)log₃(1/4) + log₃(x)
- g(x) = √((x-3)²)
- h(x) = 5x - 2x²
- r(x) = 2^(3x+1) - 2^(x-2)
- l(x) = 1/√(x)
We will explore the domain of each function and solve equations involving them, providing a comprehensive understanding of their behavior and properties.
1.1 Write down Df, the domain of f, and then solve
The domain of a function, denoted as Df, is the set of all possible input values (x-values) for which the function produces a real and defined output. Determining the domain involves identifying any restrictions on the input values, such as division by zero, square roots of negative numbers, or logarithms of non-positive numbers. For the function f(x) = (1/3)log₃(1/4) + log₃(x), we need to consider the logarithmic term. Logarithms are only defined for positive arguments. Therefore, the argument of the logarithm, which is x in this case, must be greater than zero. This restriction defines the domain of the function f(x). To find the domain, we set the argument of the logarithm to be greater than zero:
x > 0
Thus, the domain of f(x), denoted as Df, is the set of all positive real numbers. This can be expressed in interval notation as:
Df = (0, ∞)
To further illustrate this, consider the properties of logarithmic functions. The logarithm of a number represents the exponent to which the base must be raised to obtain that number. For example, log₃(9) = 2 because 3² = 9. However, logarithms are undefined for non-positive numbers because there is no exponent to which the base can be raised to obtain a non-positive result. This fundamental property of logarithms dictates the domain restriction for f(x).
Now, let's consider solving equations involving f(x). To solve an equation of the form f(x) = y, where y is a constant, we need to isolate x. This involves using the properties of logarithms to simplify the equation and express x in terms of y. For example, if we want to solve f(x) = 0, we would set the function equal to zero and proceed as follows:
(1/3)log₃(1/4) + log₃(x) = 0
To solve this equation, we first isolate the logarithmic terms. We can rewrite the equation as:
log₃(x) = -(1/3)log₃(1/4)
Using the property of logarithms that allows us to move a constant multiplier inside the logarithm as an exponent, we get:
log₃(x) = log₃((1/4)^(-1/3))
Since the logarithms are equal, their arguments must be equal as well:
x = (1/4)^(-1/3)
Simplifying the expression, we find:
x = 4^(1/3)
Thus, the solution to the equation f(x) = 0 is x = 4^(1/3), which is the cube root of 4. This solution lies within the domain of f(x), as it is a positive real number.
1.2 Solve f(x) = 0
Solving the equation f(x) = 0 involves finding the value(s) of x that make the function equal to zero. This often requires applying algebraic techniques and the properties of the functions involved. In the case of f(x) = (1/3)log₃(1/4) + log₃(x), we are dealing with a logarithmic function. To solve the equation, we will use the properties of logarithms to isolate x. As we determined earlier, the equation f(x) = 0 can be written as:
(1/3)log₃(1/4) + log₃(x) = 0
First, we isolate the logarithmic terms by moving the constant term to the right side of the equation:
log₃(x) = -(1/3)log₃(1/4)
Next, we use the power rule of logarithms, which states that logₐ(bⁿ) = nlogₐ(b), to simplify the right side of the equation. We move the -1/3 coefficient into the logarithm as an exponent:
log₃(x) = log₃((1/4)^(-1/3))
Now, we have logarithms with the same base on both sides of the equation. Since the logarithmic function is one-to-one, if logₐ(b) = logₐ(c), then b = c. Therefore, we can equate the arguments of the logarithms:
x = (1/4)^(-1/3)
To simplify the expression (1/4)^(-1/3), we first address the negative exponent. A negative exponent indicates the reciprocal of the base raised to the positive exponent. So, (1/4)^(-1/3) is the same as 4^(1/3). The exponent 1/3 represents the cube root. Therefore, we have:
x = 4^(1/3)
This means x is the cube root of 4. We can write this as:
x = ³√4
To approximate this value, we can use a calculator. The cube root of 4 is approximately 1.5874. Since this value is positive, it lies within the domain of f(x), which we determined earlier to be (0, ∞). Therefore, the solution x = ³√4 is a valid solution to the equation f(x) = 0.
In summary, solving f(x) = 0 involved using the properties of logarithms to isolate x and simplify the equation. We found that x = 4^(1/3), which is the cube root of 4. This solution is within the domain of the function f(x), making it a valid solution.
1.3 Write down the domain of g
The domain of the function g(x) = √((x-3)²) is the set of all real numbers for which the function produces a real output. In this case, we have a square root function, and we need to consider the restriction that the expression inside the square root must be non-negative. However, we also have a square inside the square root, which simplifies the analysis. The function g(x) can be rewritten as:
g(x) = √((x-3)²)
The expression (x-3)² is a square, and the square of any real number is always non-negative (i.e., greater than or equal to zero). This is because if (x-3) is positive, its square is positive, and if (x-3) is negative, its square is also positive. If (x-3) is zero, its square is zero. Therefore, (x-3)² is always non-negative for any real number x. Since the expression inside the square root is always non-negative, the square root is always defined for any real number x. Thus, there are no restrictions on the values of x for which g(x) is defined.
Another way to understand this is to recognize that the square root of a square is the absolute value. So, g(x) can also be written as:
g(x) = |x - 3|
The absolute value function is defined for all real numbers. The absolute value of a number is its distance from zero, which is always non-negative. Since the absolute value is defined for all real numbers, the function g(x) = |x - 3| is also defined for all real numbers.
Therefore, the domain of g(x), denoted as Dg, is the set of all real numbers. In interval notation, this is written as:
Dg = (-∞, ∞)
To illustrate this further, consider some examples. If x = 0, then g(0) = √((0-3)²) = √(9) = 3. If x = 3, then g(3) = √((3-3)²) = √(0) = 0. If x = 6, then g(6) = √((6-3)²) = √(9) = 3. In each case, the function g(x) produces a real output. This confirms that the domain of g(x) is indeed the set of all real numbers.
In summary, the domain of g(x) = √((x-3)²) is all real numbers because the square inside the square root ensures that the expression inside the square root is always non-negative. This makes the square root defined for all real numbers x.
1.4 Solve g(x) = 0
To solve the equation g(x) = 0, where g(x) = √((x-3)²), we need to find the value(s) of x that make the function equal to zero. As we discussed earlier, g(x) can be rewritten as the absolute value function:
g(x) = |x - 3|
So, we are solving the equation:
|x - 3| = 0
The absolute value of a number is zero if and only if the number itself is zero. Therefore, for |x - 3| to be equal to zero, the expression inside the absolute value, which is (x - 3), must be equal to zero. This gives us the equation:
x - 3 = 0
To solve for x, we add 3 to both sides of the equation:
x = 3
Thus, the solution to the equation g(x) = 0 is x = 3. This means that when x is equal to 3, the function g(x) produces an output of 0. To verify this, we can substitute x = 3 back into the original function:
g(3) = √((3-3)²) = √(0²) = √0 = 0
This confirms that x = 3 is indeed a solution to the equation g(x) = 0.
Geometrically, the function g(x) = |x - 3| represents the distance between x and 3 on the number line. The equation |x - 3| = 0 asks for the value(s) of x that are a distance of 0 from 3. There is only one such value, which is x = 3 itself.
In summary, solving g(x) = 0 involved recognizing that g(x) is equivalent to the absolute value function |x - 3|. We then used the property that the absolute value of a number is zero if and only if the number itself is zero to find the solution x = 3.
1.5 Write down the domain of h
The domain of the function h(x) = 5x - 2x² is the set of all real numbers for which the function produces a real output. The function h(x) is a polynomial function, specifically a quadratic function. Polynomial functions are defined for all real numbers, meaning there are no restrictions on the values of x that can be used as inputs. Polynomials consist of terms involving non-negative integer powers of x, and operations of addition, subtraction, and multiplication, all of which are defined for real numbers. In the case of h(x), we have two terms: 5x and -2x². The first term, 5x, is a linear term, and the second term, -2x², is a quadratic term. Both of these terms are defined for all real numbers. The operations of subtraction and multiplication are also defined for all real numbers. Therefore, the function h(x) = 5x - 2x² is defined for all real numbers.
There are no restrictions such as division by zero, square roots of negative numbers, or logarithms of non-positive numbers that would limit the domain of h(x). The domain of a polynomial function is always the set of all real numbers because we can substitute any real number for x and obtain a real number output.
Thus, the domain of h(x), denoted as Dh, is the set of all real numbers. In interval notation, this is written as:
Dh = (-∞, ∞)
To further illustrate this, consider some examples. If x = 0, then h(0) = 5(0) - 2(0)² = 0. If x = 1, then h(1) = 5(1) - 2(1)² = 5 - 2 = 3. If x = -1, then h(-1) = 5(-1) - 2(-1)² = -5 - 2 = -7. In each case, the function h(x) produces a real output. This confirms that the domain of h(x) is indeed the set of all real numbers.
In summary, the domain of h(x) = 5x - 2x² is all real numbers because h(x) is a polynomial function, and polynomial functions are defined for all real numbers. There are no restrictions on the values of x that can be used as inputs for h(x).
1.6 Solve h(x) = 0
To solve the equation h(x) = 0, where h(x) = 5x - 2x², we need to find the value(s) of x that make the function equal to zero. This involves finding the roots or zeros of the quadratic function. The equation we need to solve is:
5x - 2x² = 0
This is a quadratic equation, which can be solved by factoring, using the quadratic formula, or completing the square. In this case, factoring is the most straightforward approach. We can factor out a common factor of x from both terms:
x(5 - 2x) = 0
Now, we have a product of two factors equal to zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for x:
- x = 0
- 5 - 2x = 0
The first equation, x = 0, gives us one solution directly. For the second equation, we solve for x:
5 - 2x = 0
Add 2x to both sides:
5 = 2x
Divide both sides by 2:
x = 5/2
Thus, the solutions to the equation h(x) = 0 are x = 0 and x = 5/2. These are the x-intercepts of the graph of the quadratic function h(x).
To verify these solutions, we can substitute them back into the original equation:
For x = 0:
h(0) = 5(0) - 2(0)² = 0 - 0 = 0
For x = 5/2:
h(5/2) = 5(5/2) - 2(5/2)² = 25/2 - 2(25/4) = 25/2 - 25/2 = 0
Both solutions satisfy the equation h(x) = 0, confirming that they are correct.
In summary, solving h(x) = 0 involved factoring the quadratic expression 5x - 2x² and applying the zero-product property. We found two solutions: x = 0 and x = 5/2. These are the values of x that make the function h(x) equal to zero.
1.7 Write down the domain of r
The domain of the function r(x) = 2^(3x+1) - 2^(x-2) is the set of all real numbers for which the function produces a real output. The function r(x) involves exponential terms, which are defined for all real number exponents. Exponential functions have the form aˣ, where a is a positive constant and x is the exponent. The base a can be any positive real number, and the exponent x can be any real number. In the case of r(x), we have two exponential terms: 2^(3x+1) and 2^(x-2).
The first term, 2^(3x+1), has a base of 2 and an exponent of 3x+1. The exponent 3x+1 is a linear expression, which is defined for all real numbers x. Since the base 2 is positive, the exponential term 2^(3x+1) is defined for all real numbers x.
Similarly, the second term, 2^(x-2), has a base of 2 and an exponent of x-2. The exponent x-2 is also a linear expression, which is defined for all real numbers x. Since the base 2 is positive, the exponential term 2^(x-2) is defined for all real numbers x.
The function r(x) is the difference between these two exponential terms. The operation of subtraction is defined for all real numbers. Therefore, the function r(x) is defined for all real numbers x.
There are no restrictions such as division by zero, square roots of negative numbers, or logarithms of non-positive numbers that would limit the domain of r(x). Exponential functions are defined for all real number exponents, and the subtraction operation does not introduce any additional restrictions.
Thus, the domain of r(x), denoted as Dr, is the set of all real numbers. In interval notation, this is written as:
Dr = (-∞, ∞)
To further illustrate this, consider some examples. If x = 0, then r(0) = 2^(3(0)+1) - 2^(0-2) = 2¹ - 2⁻² = 2 - 1/4 = 7/4. If x = 1, then r(1) = 2^(3(1)+1) - 2^(1-2) = 2⁴ - 2⁻¹ = 16 - 1/2 = 31/2. If x = -1, then r(-1) = 2^(3(-1)+1) - 2^(-1-2) = 2⁻² - 2⁻³ = 1/4 - 1/8 = 1/8. In each case, the function r(x) produces a real output. This confirms that the domain of r(x) is indeed the set of all real numbers.
In summary, the domain of r(x) = 2^(3x+1) - 2^(x-2) is all real numbers because exponential functions are defined for all real number exponents, and the subtraction operation does not introduce any additional restrictions.
1.8 Solve r(x) = 0
To solve the equation r(x) = 0, where r(x) = 2^(3x+1) - 2^(x-2), we need to find the value(s) of x that make the function equal to zero. This involves solving an equation with exponential terms. The equation we need to solve is:
2^(3x+1) - 2^(x-2) = 0
To solve this equation, we first isolate the exponential terms by adding 2^(x-2) to both sides:
2^(3x+1) = 2^(x-2)
Now, we have exponential expressions with the same base (2) on both sides of the equation. If a^m = a^n, then m = n, provided that a is a positive number not equal to 1. In this case, the base is 2, which satisfies this condition. Therefore, we can equate the exponents:
3x + 1 = x - 2
This is a linear equation, which we can solve for x. Subtract x from both sides:
2x + 1 = -2
Subtract 1 from both sides:
2x = -3
Divide both sides by 2:
x = -3/2
Thus, the solution to the equation r(x) = 0 is x = -3/2. This means that when x is equal to -3/2, the function r(x) produces an output of 0.
To verify this solution, we can substitute x = -3/2 back into the original equation:
r(-3/2) = 2^(3(-3/2)+1) - 2^(-3/2-2)
Simplify the exponents:
r(-3/2) = 2^(-9/2+1) - 2^(-3/2-4/2)
r(-3/2) = 2^(-7/2) - 2^(-7/2)
r(-3/2) = 0
The solution x = -3/2 satisfies the equation r(x) = 0, confirming that it is correct.
In summary, solving r(x) = 0 involved isolating the exponential terms, equating the exponents, and solving the resulting linear equation. We found one solution: x = -3/2. This is the value of x that makes the function r(x) equal to zero.
1.9 Write down the domain of l
The domain of the function l(x) = 1/√(x) is the set of all real numbers for which the function produces a real output. This function involves both a square root and a reciprocal (division), so we need to consider the restrictions imposed by each of these operations. The square root function, √(x), is defined only for non-negative values of x. This means that x must be greater than or equal to zero. However, since the square root is in the denominator of a fraction, we must also ensure that the denominator is not zero. If the denominator is zero, the function is undefined because division by zero is not allowed.
Therefore, we have two conditions to consider:
- The expression inside the square root must be non-negative: x ≥ 0
- The denominator cannot be zero: √(x) ≠ 0
The second condition implies that x cannot be zero. If x were zero, then √(x) would be zero, and the function would be 1/0, which is undefined. Combining these two conditions, we find that x must be greater than zero, but not equal to zero. This means that x must be a positive real number.
Thus, the domain of l(x), denoted as Dl, is the set of all positive real numbers. In interval notation, this is written as:
Dl = (0, ∞)
This domain excludes zero because the square root of zero is zero, and division by zero is undefined. It also excludes negative numbers because the square root of a negative number is not a real number.
To further illustrate this, consider some examples. If x = 1, then l(1) = 1/√(1) = 1/1 = 1. If x = 4, then l(4) = 1/√(4) = 1/2. If x = 0.25, then l(0.25) = 1/√(0.25) = 1/0.5 = 2. In each case, the function l(x) produces a real output for positive values of x. However, if x = 0, then l(0) = 1/√(0) = 1/0, which is undefined. If x = -1, then l(-1) = 1/√(-1), which is not a real number.
In summary, the domain of l(x) = 1/√(x) is the set of all positive real numbers because the square root function requires a non-negative argument, and the reciprocal function requires a non-zero denominator. The combination of these restrictions limits the domain to positive real numbers.
Summary
In this exploration of functions, we determined the domains of logarithmic, square root, polynomial, exponential, and reciprocal functions. We also solved equations involving these functions, demonstrating the importance of understanding their properties and restrictions. By identifying the permissible input values and applying appropriate algebraic techniques, we can effectively analyze and manipulate various types of functions in mathematics and beyond. Understanding the domain of a function is essential for its valid and meaningful application in mathematical contexts and real-world scenarios.
Functions are the cornerstone of mathematics, providing a framework for describing relationships between quantities. Mastering the concepts of domain and equation solving is crucial for success in calculus, analysis, and various applications of mathematics.
