Complete the following. (a) Find f(x) for the indicated values of x, if possible. (b) Find the d

Complete the following. (a) Find f(x) for the indicated...

Key Concepts

Function Evaluation

This concept involves substituting specific input values into the function's expression and simplifying the result. It ensures that for given values, the function provides a meaningful output, provided that the operations (like division) are valid.

Domain of a Function

The domain of a function is the set of all possible input values for which the function is defined. In rational functions, the domain is typically restricted by excluding values that cause the denominator to be zero, as division by zero is undefined.

Rational Functions

Rational functions are expressions that involve the ratio of two polynomials or algebraic expressions. The key feature is that the denominator must not equal zero, which is critical when determining the domain of the function. This concept is essential for understanding both the evaluation of the function and its restrictions.

Transcript

00:01 Okay, so in this problem, we're given this function, f of x equals 1 over root 2 minus x, and we want to find f of x for the indicated values of x if possible.

00:11 So the indicated values are, part a, x equals 1, and x equals a plus 2.

00:20 So first, let's find f of x for x equals 1.

00:26 So we're just going to plug in x equals 1 into f of x as x.

00:29 So f of 1 equals 1 over root 2 minus 1 in for x.

00:40 All right.

00:41 So why don't we try to simplify this out to put it in a more rational form? we can multiply by root 2 plus 1 over root 2 plus 1, which is just the same as 1.

00:58 And so multiplying by one doesn't change the function in any way or the value in any way.

01:04 So we get on the top, root 2 plus 1, times 1 is just root 2 plus 1.

01:10 And then on the bottom, we do foil.

01:12 First, outer inner last.

01:14 So first, root 2 times root 2 is root 2 squared, which is just 2.

01:19 Then we have outer root 2 times 1, root 2, inner minus 1 times root 2 is negative root 2.

01:29 And last negative 1 times 1 is negative 1.

01:35 Okay, well, we can cancel these terms, and we get 2 minus 1 is 1, so just root 2 plus 1 is equal to f of 1.

01:45 So that's the value of f of x for x equals 1.

01:48 Now for f of a plus 2, we have 1 over root 2 minus a plus 2 for some arbitrary constant a.

02:06 And we have to make sure that we put these, this term in parentheses, and we're plugging in for x.

02:13 Then we distribute that negative into the parentheses...

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