Linear Approximations and Applications

Question

Use linear approximation, i.e. the tangent line, to approximate 1.7^7 as follows:

Let 𝑓(𝑥) = 𝑥^7. Find the equation of the tangent line to 𝑓 (𝑥) at 𝑥 = 2

𝐿(𝑥) =   

Using this, we find our approximation for 1.7^7 is   

Answer 1

The tangent line has equation y=128+448(x-2).

Both the tangent line and f(x) are so steep that for delta x=-.3, the approximation is not even close!

Answer 2

Given the tangent line equation \( y = 128 + 448(x - 2) \), you can see it has a slope of 448.

For a small change in \( x \), \( \Delta x = -0.3 \), the change in \( y \) using the tangent line approximation would be:

\[

\Delta y \approx \text{slope} \times \Delta x = 448 \times (-0.3) = -134.4

\]

If the approximation using the tangent line is not close for \(\Delta x = -0.3\), it suggests that the function \( f(x) \) is indeed very steep and possibly highly curved around that point. The linear approximation given by the tangent line might not accurately reflect the behavior of the function \( f(x) \) over this interval, especially if the function \( f(x) \) has a significant curvature or changes rapidly.