How to Find the Domain of Algebraic Functions

TL;DR

To find the domain of an algebraic function, identify the values that make the function undefined, such as causing division by zero or taking the square root of a negative number. For example, the domain of g(x) = 1/(√6 - |x|) is -6 < x < 6, and for h(x), it is all real numbers except x = 9 and x = -10.

Transcript

Let's do some more examples finding do mains of functions. So let's say we have a function g of x. So this is our function definition here tells us, look, if we have an input x, the output g of x is going to be equal to 1 over the square root of 6 minus -- we write this little bit neater, 1 over the square root of 6 minus the absolute value of x So... Read More

Key Insights

• 😫 The domain of a function is the set of all inputs for which the function is defined.
• 🥡 To determine the domain, consider the values of x that would result in the function being undefined, such as dividing by zero or taking the square root of a negative number.
• ☺️ The domain of g(x) = 1/(√6 - |x|) is -6 < x < 6.

Questions & Answers

Q: How do you determine the domain of a function?

The domain of a function is determined by considering the values of x that would make the function undefined, such as dividing by zero or taking the square root of a negative number.

Q: What is the domain of g(x) = 1/(√6 - |x|)?

The domain of g(x) is -6 < x < 6, as the function is undefined when 6 - |x| ≤ 0.

Q: What is the domain of h(x) = (x + 10)/(x + 10)(x - 9)(x - 5)(x - 5)?

The domain of h(x) is all real numbers except x = 9 and x = -10, as these values would result in dividing by zero.

Q: Can the function g(x) = 1/(√6 - |x|) be simplified further?

No, simplifying the function by canceling out the x + 10 terms would create a different function definition, which would be defined at x = -10. The original function is only defined for -6 < x < 6.

Summary & Key Takeaways

• The domain of a function g(x) is defined as the set of all inputs for which the function is defined.

• For g(x) = 1/(√6 - |x|), the function is defined if 6 - |x| > 0, which can be simplified to -6 < x < 6.

• For h(x) = (x + 10)/(x + 10)(x - 9)(x - 5)(x - 5), the function is undefined if x = 9 or x = -10, but it is defined for any other value of x.