Math Problem Statement

find the natural domain and range for the function f(x)=1/( square root of x + 2)

Solution

To find the natural domain and range of the function f(x)=1x+2f(x) = \frac{1}{\sqrt{x} + 2}, let's break it down into steps.

1. Domain:

The domain of a function consists of all the values of xx for which the function is defined.

For f(x)=1x+2f(x) = \frac{1}{\sqrt{x} + 2}, we have two key considerations:

  • Square root condition: The expression under the square root, xx, must be non-negative. So, x0x \geq 0.
  • Denominator condition: The denominator, x+2\sqrt{x} + 2, must not be zero. Since x0\sqrt{x} \geq 0 for all x0x \geq 0, x+2>0\sqrt{x} + 2 > 0. Therefore, there are no additional restrictions from this condition.

Thus, the natural domain of f(x)f(x) is all xx such that x0x \geq 0.

Domain: [0,)[0, \infty)

2. Range:

The range of a function consists of all possible values that f(x)f(x) can take.

To find the range of f(x)f(x):

  • As xx increases, x\sqrt{x} increases, causing x+2\sqrt{x} + 2 to increase. Since f(x)=1x+2f(x) = \frac{1}{\sqrt{x} + 2}, as x+2\sqrt{x} + 2 increases, f(x)f(x) decreases.
  • At x=0x = 0, x=0\sqrt{x} = 0, so f(0)=10+2=12f(0) = \frac{1}{0 + 2} = \frac{1}{2}.
  • As xx approaches infinity, x\sqrt{x} approaches infinity, so x+2\sqrt{x} + 2 also approaches infinity, making f(x)f(x) approach 0.

Thus, the smallest value of f(x)f(x) is 0 (approached as xx approaches infinity), and the largest value is 12\frac{1}{2} (when x=0x = 0).

Range: (0,12](0, \frac{1}{2}]

Summary

  • Domain: [0,)[0, \infty)
  • Range: (0,12](0, \frac{1}{2}]

Would you like further details or explanations?

Here are five related questions you might find helpful:

  1. How would the domain and range change if the function were f(x)=1x1+2f(x) = \frac{1}{\sqrt{x - 1} + 2}?
  2. What happens to the range if the function was f(x)=1x+1f(x) = \frac{1}{\sqrt{x} + 1} instead?
  3. How do you find the domain and range for the function g(x)=1x24g(x) = \frac{1}{\sqrt{x^2 - 4}}?
  4. Can you determine the domain and range of h(x)=12x+1+3h(x) = \frac{1}{\sqrt{2x + 1} + 3}?
  5. What if the function were f(x)=1xf(x) = \frac{1}{\sqrt{x}}? How would the domain and range differ?

Tip: Always check for conditions that could make the denominator zero or the expression under a square root negative when finding the domain of a function.

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Math Problem Analysis

Mathematical Concepts

Functions
Square roots
Domain
Range

Formulas

Function notation
Square root properties

Theorems

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Suitable Grade Level

High School