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Chapter 3: Problem 36

Evaluate the function at the indicated value of \(x\). Round your result to three decimal places. $$ \begin{aligned} &\text { Function Value }\\\ &f(x)=1.5 e^{x / 2} \quad x=240 \end{aligned} $$

Short Answer

Expert verified
The result of \(f(240)\) for the given function \(f(x)=1.5e^{x/2}\) is very large to calculate. It's practically considered as infinity.

Step by step solution

01

Substitute the value

First, substitute \(x=240\) into the given function. Afterwards the function becomes \(f(240)=1.5e^{240/2}\).
02

Calculate the exponent

Next, calculate the exponent. \(240/2 = 120\). This simplifies the function to: \[f(240)=1.5e^{120}\].
03

Apply properties of exponent

The next step is to apply properties of exponents. This is done by multiplying the coefficient of the function, which is 1.5 by the result of exponential function \(e^{120}\). Regardless, calculating \(e^{120}\) might result in a really large number or an overflown error in some calculators. So the real simplification can be seen in the next step.
04

Evaluate the function

Finally, you evaluate the function for the given value and round the result to three decimal places. Note that the exponential function value \(e^{120}\) itself yields a really large number - far bigger than the numbers we usually use. Thus, trying to calculate that might result in an overflow error. So, it is practically infeasible to have a real numerical value for \(f(240)\). Therefore, we can consider it as approximately infinitive.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a key process in mathematics where we determine the output of a function for a specific input value. In our problem, this involves substituting a value for the variable in the function equation.
When given a function like \( f(x) = 1.5e^{x/2} \), we substitute the value of \(x = 240\). This substitution changes the problem into evaluating \( f(240) = 1.5e^{120} \), automatically transforming the problem into a simpler arithmetic task.
  • The first step is to replace \(x\) with the given number.
  • Simplify the expression if necessary, before computing the final result.
A critical part of function evaluation is understanding function behavior and simplifying where possible to aid computation.
Exponential Growth
Exponential growth refers to the rapid increase that occurs according to the power of a constant base with respect to an exponent.
In our problem, we deal with the exponential function \( e^{x/2} \), a standard representation of exponential growth. The term \(e\) denotes Euler's number, an irrational and transcendental number, approximately equal to 2.718.
  • Exponential functions grow by constant multiplicative rates, unlike linear functions which grow by constant additive rates.
  • Such functions model phenomena where growth accelerates over time, such as population growth, viral spread, and compound interest.
At higher exponents, the function values grow tremendously, as we see with \( e^{120} \), which results in a number too large for standard numerical calculations. This characteristic is pivotal in understanding why some exponential expressions are approximated instead of calculated outright.
Rounding and Approximation
Rounding and approximation are essential skills in handling numerical results, especially when dealing with large numbers arising from functions like exponential growth.
When evaluating \( f(240) = 1.5e^{120} \), the result becomes impractically large, often requiring approximation.
  • Rounding is done to make numbers more manageable, typically to the nearest decimal point specified鈥攊n this case, three decimal places.
  • In scenarios where exact values are unwieldy, approximation provides a feasible method to convey the result's magnitude.
In our problem, the answer itself becomes insignificant due to overflowing limits of numerical systems, thus illustrating practical rounding as an outcome that can assume some values are effectively infinite, as seen in cases of extraordinarily large exponential products.

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Most popular questions from this chapter

Solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$\frac{1+\ln x}{2}=0$$

A laptop computer that costs $$\$ 1150$$ new has a book value of $$\$ 550$$ after 2 years. (a) Find the linear model \(V=m t+b\). (b) Find the exponential model \(V=a e^{k t}\) (c) Use a graphing utility to graph the two models in the same viewing window. Which model depreciates faster in the first 2 years? (d) Find the book values of the computer after 1 year and after 3 years using each model. (e) Explain the advantages and disadvantages of using each model to a buyer and a seller.

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Use the acidity model given by \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right],\) where acidity \((\mathrm{pH})\) is a measure of the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. Find the \(\mathrm{pH}\) if \(\left[\mathrm{H}^{+}\right]=2.3 \times 10^{-5}\).
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