Problem 51 - Precalculus with Calculus Previews

Question

Find at least one function defined implicitly by the given equation. Give the domain of each function. \((y+1)^{2}=-5(x-2)\)

Step-by-Step Solution

Answer

Functions are \(y = -1 \pm \sqrt{-5(x-2)}\), with domain \(x \leq 2\).

Step 1: Isolate y

Starting with the equation \((y+1)^2 = -5(x-2)\), we make \(y\) the subject by taking the square root of both sides:\[y+1 = \pm \sqrt{-5(x-2)}\]This gives two potential solutions for \(y\):\[y = -1 + \sqrt{-5(x-2)}\quad \text{and}\quad y = -1 - \sqrt{-5(x-2)}\]

Step 2: Check for Real Solutions

For \(y\) to be a real number, the expression under the square root, \(-5(x-2)\), must be non-negative, since we cannot take the square root of a negative number in the real number system. Set \(-5(x-2) \geq 0\):\[-5(x-2) \geq 0\]Solving gives \(x-2 \leq 0\), or \(x \leq 2\).

Step 3: Determine Domain

Since \(x-2 \leq 0\), it follows that the domain for \(x\) is restricted to values where \(x \leq 2\). This applies to both potential solutions for \(y\). Therefore, the domain for each function is \(x \leq 2\).

Key Concepts

Square Root

In mathematics, the square root of a number is one of the two identical factors that, when multiplied together, produce that number. For example, the square root of 9 is 3 because 3 multiplied by 3 equals 9.
The square root symbol is represented as \( \sqrt{} \), and finding the square root of a number is called "taking the square root." It's important to note that every positive number has both a positive and a negative square root. For example, both 3 and -3 are square roots of 9. This is because \( (-3) \times (-3) = 9 \).
When we apply this to functions, especially implicit ones as seen in the example, taking the square root can help solve for variables. However, we need to ensure that the expression inside the square root is non-negative because, within the real numbers, we cannot find the square root of a negative number.

Real Numbers

Real numbers are a set of numbers that include all the numbers you can think of, except for the imaginary numbers. This set includes positive numbers, negative numbers, zero, and fractional numbers.
When dealing with square roots, as previously mentioned, real numbers are critical: we can only take the square root of a non-negative number and yield a real number result. In other words, if the expression inside a square root is negative, the result is not a real number.
The rule that the square root of a non-negative number results in a real number is crucial when finding functions derived from an equation. As we rearrange and solve equations, we continuously check that our solutions remain within the real number domain.

Domain of a Function

The domain of a function is all the possible input values (often \( x \)) that allow the function to work without restrictions or errors. Simply put, it's all the numbers you can use in a function that will not lead to math errors, like dividing by zero or square rooting a negative number in the real number system.
For example, in the exercise given with the implicit equation \((y+1)^2 = -5(x-2)\), the domain has to consider the condition when the square root is defined. In this case, we derived that \(-5(x-2) \geq 0\) must hold true for real numbers under the square root. Solving this inequality gives \(x \leq 2\), which sets the domain of the function.
Understanding the domain ensures that we don't use values for \( x \) that make the function impossible or undefined. This basic check is essential in ensuring that our function behaves predictably within real numbers.