Question

Consider the function h(x)=18x . Given is a table with x-values that get closer to  e≈2.71828182 (Euler’s number) and their corresponding h(x)  values. Use the sequence in the table to approximate 18e  correct to three decimal places.

Answer 1

To approximate \( 18e \) using the function \( h(x) = 18x \) and considering the values of \( x \) that approach \( e \), we can evaluate \( h(x) \) at values close to \( e \).

Since you have a table of values, you would normally plug in each of these \( x \) values and calculate \( h(x) = 18x \). For example, if the x-values in your table are close to \( e \) such as \( 2.7, 2.71, 2.718, 2.7182, \) etc., you would find:

- For \( x = 2.7 \), \( h(2.7) = 18 \times 2.7 = 48.6 \)
- For \( x = 2.71 \), \( h(2.71) = 18 \times 2.71 = 48.78 \)
- For \( x = 2.718 \), \( h(2.718) = 18 \times 2.718 = 48.924 \)
- For \( x = 2.7182 \), \( h(2.7182) = 18 \times 2.7182 = 48.9256 \)

As you evaluate \( h(x) \) as \( x \) approaches \( e \), you will note that the values of \( h(x) \) will converge towards \( 18e \).

Now, if you take the closest value from your table (say \( x = 2.718 \)), you will find that:

\[
h(2.718) \approx 48.924
\]

Thus, rounding this result to three decimal places gives:

\[
18e \approx 48.924
\]

If you have other x-values closer to e (perhaps additional digits), you would use those to refine your approximation of \( 18e \). However, based on the example shown, \( h(2.718) \) already gives a sufficiently accurate approximation to three decimal places.