In exercises 9-12, use linear interpolation to estimate the...

In exercises 9-12, use linear interpolation to estimate the desired quantity. A company estimates that $f(x)$ thousand software games can be sold at the price of $\$ x$ as given in the table. $$\begin{array}{|l|l|l|l|}\hline x & 20 & 30 & 40 \\\hline f(x) & 18 & 14 & 12 \\\hline\end{array}$$

In exercises 9-12, use linear interpolation to estimate the desired quantity.
A company estimates that $f(x)$ thousand software games can be sold at the price of $\$ x$ as given in the table.
$$\begin{array}{|l|l|l|l|}\hline x & 20 & 30 & 40 \\\hline f(x) & 18 & 14 & 12 \\\hline\end{array}$$

Key Concepts

Linear Interpolation

Linear interpolation is a method used to estimate an unknown value within two known data points by assuming that the change between these points is linear. This technique uses the linear relationship between the neighboring data values to construct a straight line, along which any intermediate value can be approximated.

Interpolation Formula

The interpolation formula provides a mathematical approach to compute the estimated value. Typically, it is written as y = y? + ((x - x?) / (x? - x?)) · (y? - y?), where (x?, y?) and (x?, y?) are the closest known points. This formula calculates the proportion of the difference between the known function values corresponding to the relative position of x within the interval.

Tabulated Data

Tabulated data refers to a set of known data points presented in a table format. Such tables often depict pairs of independent and dependent values, which are used as the basis for various estimation techniques, including linear interpolation, to determine values that were not directly measured or recorded.

Estimation

Estimation in this context involves predicting or approximating an unknown value based on available data. By assuming a linear progression between two known points, estimation allows the calculation of an intermediate value, which is useful in many practical applications where continuous data is not fully available.

Transcript

00:01 Okay, so for this problem, the two points in the table that we're going to use are 2018 and 30 .14.

00:19 So we're going to use these two.

00:21 So first what i want to do is i want to find the slope of the point or the line that goes through these two points.

00:30 So to do that, i'm going to do the difference of the y coordinates over the difference of the x coordinates.

00:36 That's going to be 14 minus 18 over 30 minus 20.

00:43 And if we simplify this, we will get a slope of negative 2 over 5.

00:50 Okay, so what i want to do is i want to make a linear approximation formula.

00:56 First we're going to start with point slope form in order to do that.

01:00 So we're going to do y minus.

01:02 So our y coordinate, you can pick any of these two points, but i'm going to use the first because if you notice the x coordinate in the formula is going to be closer to our desired value.

01:15 So i'm going to use this one as the y.

01:17 So i'm going to do y minus 18 is equal to m, which is going to be this right here, which we just calculated as the slope, and then x minus this x coordinate here.

01:29 So x minus 20.

01:31 So here i'm just going to solve for y.

01:33 So y is going to be negative two -fifths times x minus 20.

01:39 Plus 18.

01:43 So this is going to be actually our l of x.

01:47 Now what we need to find is l of our desired quantity, 24.

01:53 So l of 24, just have to plug it right into here...