Problem 14 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

If \(f(x)=\sqrt{\frac{1}{x+1}}\), find the following. Simplify your answer where possible. (a) \(f(0)\) (b) \(f(3)\) (c) \(f\left(-\frac{1}{4}\right)\) (d) \(f(b)\) (e) \(f(b-1)\) (f) \(f(b+3)\) (g) \([f(7)]^{2}\) (h) \(f\left(b^{2}\right)\) (i) \([f(b)]^{2}\)

Step-by-Step Solution

Answer

(a) 1 \n (b) 1/2 \n (c) \(2\sqrt{3}/3\) \n (d) \(\sqrt{1/(b+1)}\) \n (e) \(\sqrt{1/b}\) \n (f) \(\sqrt{1/(b+4)}\) \n (g) 1/4 \n (h) \(\sqrt{1/(b^{2} + 1)}\) \n (i) \(1/(b+1)\)

Step 1: Substitute x = 0

Plug x = 0 into the function: \(f(0)=\sqrt{\frac{1}{0+1}}=\sqrt{1}=1\)

Step 2: Substitute x = 3

Plug x = 3 into the function: \(f(3)=\sqrt{\frac{1}{3+1}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

Step 3: Substitute x = -1/4

Plug x = -1/4 into the function: \(f\left(-\frac{1}{4}\right)=\sqrt{\frac{1}{-\frac{1}{4}+1}}=\sqrt{\frac{1}{\frac{3}{4}}}= \sqrt{\frac{4}{3}}= \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\) (multiplied by \(\sqrt{3} / \sqrt{3}\) to rationalize the denominator)

Step 4: Substitute x = b

Plug x = b into the function: \(f(b)=\sqrt{\frac{1}{b+1}}\) (can't be simplified any further)

Step 5: Substitute x = b-1

Plug x = b-1 into the function: \(f(b-1)=\sqrt{\frac{1}{(b-1)+1}}=\sqrt{\frac{1}{b}}\) (can't be simplified any further)

Step 6: Substitute x = b+3

Plug x = b+3 into the function: \(f(b+3)=\sqrt{\frac{1}{(b+3)+1}}=\sqrt{\frac{1}{b+4}}\) (can't be simplified any further)

Step 7: Substitute x = 7 and square the result

Plug x = 7 into the function and square the result: \([f(7)]^{2}=[\sqrt{\frac{1}{7+1}}]^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}\)

Step 8: Substitute x = \(b^{2}\)

Plug x = \(b^{2}\) into the function: \(f\left(b^{2}\right)=\sqrt{\frac{1}{(b^{2})+1}}\) (can't be simplified any further)

Step 9: Substitute x = b and square the result

Plug x = b into the function and square the result: \([f(b)]^{2}=[\sqrt{\frac{1}{b+1}}]^{2}=\frac{1}{b+1}\)

Key Concepts

Square Root Functions

A square root function is a type of function that involves a square root. For example, the function \( f(x)=\sqrt{\frac{1}{x+1}} \) includes a square root over a fraction. These types of functions often show up in calculus and help us understand the rate of growth, limits, and other aspects of different expressions. The radical symbol, \( \sqrt{} \), represents the square root. It indicates a number that, when multiplied by itself, gives the original number inside the radical. Understanding square roots is fundamental to working with these functions as it helps simplify expressions and analyze data efficiently.
In calculus, square root functions are especially useful for modeling situations where quantities grow at diminishing rates, such as diminishing returns or certain types of decay.

Function Evaluation

Function evaluation involves substituting a specific value into a given function. When working with \( f(x)=\sqrt{\frac{1}{x+1}} \), you simply replace \( x \) with the given number.

• For \( f(0) \), substitute \( x=0 \) which becomes \( f(0)=\sqrt{1}=1 \).
• For \( f(3) \), substitute \( x=3 \) giving \( f(3)=\sqrt{\frac{1}{4}}=\frac{1}{2} \).
• For \( f\left(-\frac{1}{4}\right) \), substitute \( x=-\frac{1}{4} \) to compute \( \sqrt{\frac{4}{3}} \).

By plugging in various values of \( x \), you find specific outputs or results for the function. This is essential for graphing, understanding function behavior, and solving real-world problems.
Evaluating functions accurately requires careful substitution and simplification to ensure correct results.

Rationalizing Denominators

Rationalizing the denominator means converting a fraction so that the denominator no longer contains a square root. This process helps make expressions easier to work with and can often be more suitable for further mathematical operations.
For instance, in the solution \( f\left(-\frac{1}{4}\right)=\sqrt{\frac{4}{3}} \) becomes \( \frac{2}{\sqrt{3}} \). Here, multiplying by \( \frac{\sqrt{3}}{\sqrt{3}} \) transforms it into \( \frac{2\sqrt{3}}{3} \).

• This step keeps fractions in a more conventional form.
• It simplifies expressions, often leading to clearer results.

Rationalizing is a standard technique in mathematics and is particularly important when working with square roots and radicals. It ensures presentations of solutions conform to mathematical conventions.

Substitution in Functions

Substituting in functions involves replacing the variable \( x \) with another quantity or expression. This is important in calculus for evaluating the function across different scenarios. For example:

• By replacing \( x \) with \( b \) in \( f(b)=\sqrt{\frac{1}{b+1}} \), it remains in a form that represents the relationship for any variable \( b \).
• Substituting \( b-1 \), \( b+3 \), or even \( b^2 \) enables us to explore how the function behaves under different conditions.

This technique allows us to handle variables that are not just numbers but entire expressions, providing flexibility. Sub substitution helps track dependencies and analyze function changes with varying inputs. It's crucial for thorough understanding and manipulation of mathematical models and real-world scenarios alike.