Problem 61 - Algebra and Trigonometry with Analytic Geometry

Question

If \(f(x)=\sqrt{x}-1\) and \(g(x)=x^{3}+1\), approximate \((f \circ g)(0.0001)\). In order to avoid calculating a zero value for \((f \circ g)(0.0001)\), rewrite the formula for \(f \circ g\) as $$ \frac{x^{3}}{\sqrt{x^{3}+1}+1} \text {. } $$

Step-by-Step Solution

Answer

\((f \circ g)(0.0001) \approx 5 \times 10^{-13}.\)

Step 1: Understanding the Functions

We are given two functions: \( f(x) = \sqrt{x} - 1 \) and \( g(x) = x^3 + 1 \). The composition \((f \circ g)(x)\) means \(f(g(x))\). Our task is to find \((f \circ g)(0.0001)\).

Step 2: Composition of Functions

First, we substitute \(g(x)\) into \(f(x)\). This means we calculate \(f(g(x)) = f(x^3 + 1) = \sqrt{x^3 + 1} - 1\).

Step 3: Rewriting the Composition

To avoid calculating a zero value, the composition \(\sqrt{x^3 + 1} - 1\) is rewritten as \(\frac{x^3}{\sqrt{x^3 + 1} + 1}\). This is a rational expression derived from multiplying numerator and denominator by the conjugate \(\sqrt{x^3 + 1} + 1\) to rationalize a form \(a - b\).

Step 4: Evaluate the Expression

Now evaluate \(\frac{x^3}{\sqrt{x^3 + 1} + 1}\) at \(x = 0.0001\). We substitute \(x = 0.0001\) into the expression: \[\frac{(0.0001)^3}{\sqrt{(0.0001)^3 + 1} + 1}\].

Step 5: Approximate the Expression

Calculate \((0.0001)^3 = 0.000000000001\). Now find \(\sqrt{0.000000000001 + 1} + 1\), which is approximately \(2\) because \(\sqrt{1\text{.000000000001}}\) is nearly \(1\).

Step 6: Final Calculation

Divide the small numerator by the approximate denominator: \[\frac{0.000000000001}{2} = 0.0000000000005.\] Hence, the approximate value of \((f \circ g)(0.0001)\) is \(5 \times 10^{-13}\).

Key Concepts

Rational Expressions

Rational expressions represent the ratio of two polynomials or algebraic expressions. In mathematics, the form \( \frac{a}{b} \) is a rational expression if both \(a\) and \(b\) are polynomials, and \(b\) is not zero. This is similar to fractions but involves variables.In the provided exercise, we encounter a rational expression while solving for the composition \((f \circ g)(x)\). Originally, the expression \(\sqrt{x^3 + 1} - 1\) needed to be evaluated. But to avoid dealing with an expression that could be close to zero and complicated square roots directly, we change it to a rational expression:

• The reformed expression is \(\frac{x^3}{\sqrt{x^3 + 1} + 1}\).
• We use the conjugate \(\sqrt{x^3 + 1} + 1\) to simplify and rationalize the original form. This technique helps in preventing division by a number that approaches zero, ensuring accurate approximations.

This modified structure makes it more manageable, especially for making calculations with very small values in the numerators.

Function Evaluation

Function evaluation involves replacing each occurrence of a variable in a function with a given number, allowing us to compute its result. This is a key concept in understanding compositions and transformations of functions.To evaluate \((f \circ g)(0.0001)\), we need to perform the following steps:

• First, evaluate the inner function, \(g(x) = x^3 + 1\), at \(x = 0.0001\).
• Next, use this result in the outer function \(f(x) = \sqrt{x} - 1\).

This nesting of functions is represented as \(f(g(x))\), ensuring we reflect every step accurately. In practice, rather than dealing with complicated forms, rewriting functions to simpler or equivalent forms aids in straightforward evaluations, especially when small or peculiar values, like 0.0001, are involved.

Mathematical Approximation

In mathematics, approximation is the process of finding a value that is close to but not exactly equal to the actual or needed number. It's particularly useful when dealing with small values or complex expressions that are difficult to compute directly.In this exercise, after substituting the small value \(x = 0.0001\) into our rational expression, we need to approximate the result:

• Calculating \((0.0001)^3\) yields a very small number, \(0.000000000001\).
• For \(\sqrt{(0.0001)^3 + 1} + 1\), the value inside the square root approaches 1, making the entire term close to 2.

We use these approximate values to perform simpler arithmetic, reducing \(\frac{0.000000000001}{2}\) to a manageable result of \(0.0000000000005\). This streamlined computation is a classic example of applying approximation to obtain practical solutions where precision may limit direct arithmetic.