Problem 62 - Algebra and Trigonometry with Analytic Geometry

Question

If \(f(x)=\frac{x^{3}}{x^{2}+x+2}\) and \(g(x)=\left(\sqrt{3 x}-x^{3}\right)^{3 / 2}\), approximate $$ \frac{(f+g)(1.12)-(f / g)(1.12)}{[(f \circ f)(5.2)]^{2}} $$

Step-by-Step Solution

Answer

Approximate value is -0.061.

Step 1: Calculate \(f(1.12)\)

First, we need to calculate \(f(1.12)\). Substitute \(x = 1.12\) into the function \(f(x) = \frac{x^3}{x^2 + x + 2}\). This gives us: \[ f(1.12) = \frac{(1.12)^3}{(1.12)^2 + 1.12 + 2}. \]Compute: \(1.12^3 = 1.404928\) and \((1.12)^2 + 1.12 + 2 = 1.2544 + 1.12 + 2 = 4.3744\).Thus, \[ f(1.12) = \frac{1.404928}{4.3744} \approx 0.321\].

Step 2: Calculate \(g(1.12)\)

Next, we find \(g(1.12)\). Substitute \(x = 1.12\) into the function \(g(x) = (\sqrt{3x} - x^3)^{3/2}\). This gives us:\[ g(1.12) = \left(\sqrt{3 \times 1.12} - (1.12)^3\right)^{3/2}. \]Compute:\(\sqrt{3 \times 1.12} = \sqrt{3.36} \approx 1.833\),so:\(1.833 - 1.404928 \approx 0.428072\).Thus, \[ g(1.12) = (0.428072)^{3/2} \approx 0.279 \].

Step 3: Compute \((f+g)(1.12)\)

Now calculate \((f+g)(1.12)\) using the values obtained:\[ (f+g)(1.12) = f(1.12) + g(1.12) = 0.321 + 0.279 = 0.600. \]

Step 4: Compute \((f/g)(1.12)\)

Next, calculate \((f/g)(1.12)\):\[ \frac{f}{g}(1.12) = \frac{f(1.12)}{g(1.12)} = \frac{0.321}{0.279} \approx 1.150. \]

Step 5: Express the Numerator

Substitute the found values into the expression for the numerator:\[ (f+g)(1.12) - \frac{f}{g}(1.12) = 0.600 - 1.150 = -0.550. \]

Step 6: Calculate \(f(f(5.2))\)

Compute \(f(5.2)\) first: \[ f(5.2) = \frac{(5.2)^3}{(5.2)^2 + 5.2 + 2}. \]Calculations:\(5.2^3 = 140.608\) and \(5.2^2 + 5.2 + 2 = 27.04 + 5.2 + 2 = 34.24\).Thus:\[ f(5.2) = \frac{140.608}{34.24} \approx 4.106. \]Calculate \(f(f(5.2))\): Substituting again using \(f(x)\):\[ f(4.106) = \frac{(4.106)^3}{(4.106)^2 + 4.106 + 2}. \]Further calculations:\(4.106^3 = 69.258\), \(4.106^2 = 16.875\). Thus,\(16.875 + 4.106 + 2 = 22.981\), therefore \[ f(4.106) = \frac{69.258}{22.981} \approx 3.014. \]

Step 7: Square \((f \, \circ \, f)(5.2)\)

Square the result from Step 6: \[ ((f \, \circ \, f)(5.2))^2 = (3.014)^2 \approx 9.084. \]

Step 8: Compute Final Expression

Substitute the numerator and the squared value from step 7 into the original expression:\[ \frac{(f+g)(1.12) - \frac{f}{g}(1.12)}{[(f \, \circ \, f)(5.2)]^2} = \frac{-0.550}{9.084} \approx -0.061. \]

Key Concepts

Function Composition

Function Composition involves creating a new function by applying one function to the results of another. When you see something like \((f \circ g)(x)\), it means you first apply \(g\) to \(x\), then apply \(f\) to the result.

This concept is crucial in mathematics because it allows us to combine simple functions into more complex ones. In practical terms, composing functions helps us gain insights into how changes in one variable affect the system represented by the function.

Consider our exercise where we compute \(f(f(5.2))\). This involves first calculating \(f(5.2)\) and then using this result as the input for another \(f\) function. Here, composition shows how a function can affect data when applied repeatedly.

Rational Functions

Rational Functions are expressions defined as the ratio of two polynomials, much like fractions but involving variables. They can be written as \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\).

In the given exercise, \(f(x) = \frac{x^3}{x^2 + x + 2}\) is a rational function. These functions are vital in algebra due to their complex behavior, including vertical asymptotes and horizontal asymptotes, which are key in understanding their graphs.

The complexity arises because, unlike linear functions, rational functions can have holes or asymptotes where the function is undefined. This means they require careful evaluation when trying to compute values, especially when dealing with divisions, as seen in \((f/g)(1.12)\).

Numerical Approximation

Numerical Approximation involves estimating the values of functions and expressions when an exact form isn't practical or necessary. This is particularly useful when dealing with complex functions where an analytic solution is not possible or is difficult to obtain.

For instance, in our problem, after calculating \(f(1.12)\) and \(g(1.12)\), we approximate these results to make further calculations manageable. By rounding figures to a reasonable number of significant digits, we ensure computations can be performed efficiently while remaining accurate enough.

This method is incredibly useful in real-world applications where time or computational power is limited. Numerical approximation allows us to get a close estimate, often enough to make critical decisions based on these approximations.

Function Evaluation

Function Evaluation is the process of finding the output of a function for a particular input value. This is one of the fundamental steps in using functions to model real-world scenarios.

In the exercise provided, evaluating functions such as \(f(1.12)\) and \(g(1.12)\) entails plugging the given \(x\)-value into the expressions for \(f\) and \(g\), respectively.

This task forms the basis for more advanced operations, such as computing new functions through addition, division, or composition, as demonstrated in our example.

The importance of mastering function evaluation lies in its ability to help us analyze the behavior of functions and predict outcomes efficiently.