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Chapter 3: Problem 10

Evaluate the function at the indicated value of \(x\). Round your result to three decimal places. $$ \begin{aligned} &\text { Function } \quad \text { Valu }\\\ &f(x)=\left(\frac{2}{3}\right)^{5 x} \quad x=\frac{3}{10} \end{aligned} $$

Short Answer

Expert verified
The value of the function \(f(x)\) at \(x=\frac{3}{10}\) is approximately 0.577.

Step by step solution

01

Substitute the value of x into the function

Replace \(x\) by the given value \(\frac{3}{10}\) in the function \(f(x)=(\frac{2}{3})^{5x}\). Thus, it becomes \(f(\frac{3}{10})=(\frac{2}{3})^{5 \times \frac{3}{10}}\).
02

Simplify the exponent

Perform the multiplication in the exponent, \(5 \times \frac{3}{10}=1.5\), to simplify the function as \(f(\frac{3}{10})=(\frac{2}{3})^{1.5}\).
03

Calculate the value of the function

The final step is to calculate the value of \((\frac{2}{3})^{1.5}\). Upon calculating, the approximate value is 0.577.

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Key Concepts

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

Function Evaluation in Exponential Functions
When dealing with functions, function evaluation simply means replacing the variable with a given value and solving the equation. Here, our function is exponential: \( f(x) = \left(\frac{2}{3}\right)^{5x} \). We need to evaluate it at \( x = \frac{3}{10} \). This process involves:
  • Substitute the variable \( x \) with \( \frac{3}{10} \) in the function.
  • This turns the equation into \( f\left(\frac{3}{10}\right) = \left(\frac{2}{3}\right)^{5 \times \frac{3}{10}} \).
We're essentially finding the value of the function for this specific input value. It鈥檚 a straightforward substitution process. Next, we simplify the exponent to continue evaluating it.
Understanding Exponents in Exponential Functions
Exponents in mathematics are a way of expressing how many times a number, known as the base, is multiplied by itself. In our function, we have an exponent presented as \( \left(\frac{2}{3}\right)^{5x} \), where \( 5x \) is the exponent.This part involves multiplying 5 by our substituted value of \( x = \frac{3}{10} \):
  • Perform the multiplication: \( 5 \times \frac{3}{10} = 1.5 \).
  • Now the exponent is simplified to 1.5, resulting in the expression \( \left(\frac{2}{3}\right)^{1.5} \).
Exponents like 1.5, which are not integers, are perfectly valid and are known as fractional exponents. Solving this involves calculating what is often called the root; here, a mix of a square root and a cube, but simplified using calculators.
Rounding the Result
Rounding means adjusting the decimal point of a number to make it simpler and easier to use or present. In mathematics problems like ours, it's crucial to round numbers to a specific decimal place. Here, the instruction is to round to three decimal places.Once the value of the function is computed, we arrive with a decimal answer. Let's say you find the result of \( \left(\frac{2}{3}\right)^{1.5} \) using a calculator:
  • You might get a number like 0.576898.
  • Rounding this to three decimal places gives us 0.577.
When rounding, observe the fourth decimal point digit: if it's 5 or more, round up the third digit; if less, keep it the same. Here, the calculation tells us to keep things precise, and for this problem, the rounded result is 0.577.

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Most popular questions from this chapter

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln x-\ln (x+1)=2$$

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$10-4 \ln (x-2)=0$$

Rewrite each verbal statement as an equation. Then decide whether the statement is true or false. Justify your answer. The logarithm of the quotient of two numbers is equal to the difference of the logarithms of the numbers.

Automobiles are designed with crumple zones that help protect their occupants in crashes. The crumple zones allow the occupants to move short distances when the automobiles come to abrupt stops. The greater the distance moved, the fewer g's the crash victims experience. (One \(g\) is equal to the acceleration due to gravity. For very short periods of time, humans have withstood as much as 40 g's.) In crash tests with vehicles moving at 90 kilometers per hour, analysts measured the numbers of g's experienced during deceleration by crash dummies that were permitted to move \(x\) meters during impact. The data are shown in the table. A model for the data is given by \(y=-3.00+11.88 \ln x+(36.94 / x),\) where \(y\) is the number of g's. $$ \begin{array}{|c|c|} \hline x & \text { g's } \\ \hline 0.2 & 158 \\ 0.4 & 80 \\ 0.6 & 53 \\ 0.8 & 40 \\ 1.0 & 32 \\ \hline \end{array} $$ (a) Complete the table using the model. $$ \begin{array}{|l|l|l|l|l|l|} \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline y & & & & & \\ \hline \end{array} $$ (b) Use a graphing utility to graph the data points and the model in the same viewing window. How do they compare? (c) Use the model to estimate the distance traveled during impact if the passenger deceleration must not exceed \(30 \mathrm{~g}\) 's. (d) Do you think it is practical to lower the number of g's experienced during impact to fewer than \(23 ?\) Explain your reasoning.

The table shows the time \(t\) (in seconds) required for a car to attain a speed of \(s\) miles per hour from a standing start. $$ \begin{array}{|c|c|} \hline \text { Speed, } s & \text { Time, } t \\ \hline 30 & 3.4 \\ 40 & 5.0 \\ 50 & 7.0 \\ 60 & 9.3 \\ 70 & 12.0 \\ 80 & 15.8 \\ 90 & 20.0 \\ \hline \end{array} $$ Two models for these data are as follows. \(t_{1}=40.757+0.556 s-15.817 \ln s\) \(t_{2}=1.2259+0.0023 s^{2}\) (a) Use the regression feature of a graphing utility to find a linear model \(t_{3}\) and an exponential model \(t_{4}\) for the data. (b) Use a graphing utility to graph the data and each model in the same viewing window. (c) Create a table comparing the data with estimates obtained from each model. (d) Use the results of part (c) to find the sum of the absolute values of the differences between the data and the estimated values given by each model. Based on the four sums, which model do you think best fits the data? Explain.
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