To estimate $e^{-0.015}$ using linearization, we follow these steps:

Step 1: Define the function

We are given the function $f(x) = e^x$.

Step 2: Calculate the linearization at $a = 0$

Linearization is an approximation of a function near a given point $a$, using the tangent line at that point. The formula for the linearization $L(x)$ of $f(x)$ at $a = 0$ is:

$L(x) = f(a) + f'(a)(x - a)$

Since $a = 0$, this becomes:

$L(x) = f(0) + f'(0)(x - 0)$

Step 3: Calculate $f(0)$ and $f'(0)$

• $f(0) = e^0 = 1$
• $f'(x) = e^x$, so $f'(0) = e^0 = 1$

Thus, the linearization at $a = 0$ is:

$L(x) = 1 + 1(x - 0) = 1 + x$

Step 4: Approximate $e^{-0.015}$

Using the linearization $L(x) = 1 + x$, we can approximate $e^{-0.015}$ by substituting $x = -0.015$:

$e^{-0.015} \approx L(-0.015) = 1 + (-0.015) = 0.985$

Thus, the linear approximation of $e^{-0.015}$ is approximately $0.985$.

Would you like further details on this process or have any other questions?

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Here are 5 related questions to expand on this topic:

1. How does linearization differ from higher-order approximations like Taylor series?
2. How accurate is this linearization approximation for small values of $x$?
3. What happens if we use a different point $a$ for linearization?
4. How would we estimate $e^{x}$ using a second-degree approximation?
5. Can linearization be applied to other types of functions, like trigonometric or logarithmic?

Tip: Linear approximations are most accurate near the point of tangency. For more precision, higher-order terms (from Taylor series) can be included.