Question 3.1.4: Using a Linear Approximation to Perform Linear Interpolation...
Using a Linear Approximation to Perform Linear Interpolation
Suppose that based on market research, a company estimates that f(x) thousand small cameras can be sold at the price of $x, as given in the accompanying table. Estimate the number of cameras that can be sold at $7.
The closest x-value to x = 7 in the table is x = 6. [In other words, this is the closest value of x at which we know the value of f(x).] The linear approximation of f(x) at x = 6 would look like
L(x)=f(6)+f^{\prime}(6)(x-6).
From the table, we know that f(6) = 84, but we do not know f'(6). Further, we can’t compute f'(x), since we don’t have a formula for f(x). The best we can do with the given data is to approximate the derivative by
f^{\prime}(6) \approx \frac{f(10)-f(6)}{10-6}=\frac{60-84}{4}=-6.
The linear approximation is then
L(x) \approx 84-6(x-6).
An estimate of the number of cameras sold at x = 7 would then be L(7) ≈ 84 − 6 = 78 thousand. We show a graphical interpretation of this in Figure 3.6, where the straight line is the linear approximation (in this case, the secant line joining the first two data points).
