Problem 13 - Calculus with Concepts in Calculus

Question

Find the numerical value of the function at the given values of \(a\). $$ g(x)=\frac{100.24 x^{6}}{(0.24) x^{6}-1} ; a=0.5,-7.31 $$

Step-by-Step Solution

Answer

\(g(0.5) \approx -1.5708\), \(g(-7.31) \approx 417.748\)

Step 1: Understand the Function

The function given is \[ g(x) = \frac{100.24 x^{6}}{(0.24)x^{6} - 1} \]Our task is to find the values of this function at specific values of \(a\), namely \(a = 0.5\) and \(a = -7.31\).

Step 2: Substitute and Simplify for a = 0.5

Substitute \(x = 0.5\) into the function \[g(0.5) = \frac{100.24 \times (0.5)^{6}}{(0.24) \times (0.5)^{6} - 1}\]Calculate \((0.5)^6 = 0.015625\). Substitute this value back into the equation:\[g(0.5) = \frac{100.24 \times 0.015625}{(0.24 \times 0.015625) - 1}\]Simplify the numerator and the denominator to evaluate \(g(0.5)\).

Step 3: Calculate g(0.5)

Calculate the numerator: \[100.24 \times 0.015625 = 1.565375\]Calculate the denominator:\[0.24 \times 0.015625 = 0.00375\]Subtracting 1 from the result of the denominator:\[0.00375 - 1 = -0.99625\]Calculate:\[g(0.5) = \frac{1.565375}{-0.99625} \approx -1.5708\]

Step 4: Substitute and Simplify for a = -7.31

Substitute \(x = -7.31\) into the function:\[g(-7.31) = \frac{100.24 \times (-7.31)^{6}}{(0.24) \times (-7.31)^{6} - 1}\]Calculate \((-7.31)^6\). Substitute that result back into the equation to evaluate \(g(-7.31)\).

Step 5: Calculate (-7.31)^6

Calculate the sixth power of \(-7.31\): \[(-7.31)^6 = 1607646.243 \]

Step 6: Calculate g(-7.31)

Use the value from step 5 to substitute into the function:Calculate the numerator:\[100.24 \times 1607646.243 = 161148084.73252 \]Calculate the denominator:\[0.24 \times 1607646.243 = 385835.09832 \]Subtract 1:\[385835.09832 - 1 = 385834.09832 \]Calculate:\[g(-7.31) = \frac{161148084.73252}{385834.09832} \approx 417.748\]

Step 7: Conclusion and Values

The calculated numerical value of the function for \(a = 0.5\) is approximately \(-1.5708\), and for \(a = -7.31\) is approximately \(417.748\).

Key Concepts

Numerical Evaluation

Numerical evaluation is a key aspect in calculus, especially when dealing with complex functions. The primary goal is to find the specific value of a function for certain inputs. This is achieved by substituting the input value into the function, simplifying the expression, and then calculating the result.
For example, in the provided exercise, we're tasked to find the outcomes for the function \( g(x) = \frac{100.24 x^{6}}{(0.24) x^{6}-1} \) at specific points \( a = 0.5 \) and \( a = -7.31 \).
Numerical evaluation involves multiple steps including substitution and actual computation where you'll deal with powers, multiplication, and subtraction. It's crucial to follow each step carefully to get accurate results. Mistakes often come from minor calculation errors, so double-checking your work can make a huge difference.

Polynomial Function

A polynomial function is an expression consisting of variables and coefficients, involving operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
The function \( g(x) = \frac{100.24 x^{6}}{(0.24) x^{6} - 1} \) involves a polynomial in the form of a ratio, or a rational function, involving the sixth power of \( x \).
This means you will deal with a fairly large increase in values as \( x \) increases, due to the exponent of 6. It is important to recognize and properly manage these polynomial terms as you evaluate the function.
Polynomial functions are fundamental in calculus due to their smooth, continuous nature allowing for straightforward analysis, making them easier to derive and integrate compared to more complex functions.

Evaluation at Specific Points

Evaluating a function at specific points means plugging in certain values for \( x \) in the function to find the output value. This is a common task in calculus that helps understand how a function behaves at certain moments.

• First, substitute the specified value into the function. For instance, if \( a = 0.5 \), replace \( x \) with 0.5 in the function \( g(0.5) = \frac{100.24 \, (0.5)^{6}}{(0.24) \, (0.5)^{6} - 1} \).
• Next, carry out the arithmetic involving powers, multiplications, and subtractions.

This process enables you to determine specific function values seen in real-life applications, where understanding how values change at different points is crucial.
Practical scenarios include physics where certain positions (or values of \( x \)) of an object are evaluated over time, impacting design choices and predictions.