To estimate $(1.999)^4$ using a linear approximation (differentials), we can follow these steps:

Steps:

1. Choose a function and point of approximation: The function is $f(x) = x^4$. We'll approximate near $x = 2$, since $1.999$ is very close to $2$.

2. Find $f'(x)$ (the derivative): $f'(x) = 4x^3$

3. Set up the linear approximation formula: The linear approximation at $x = a$ is: $f(x) \approx f(a) + f'(a)(x - a)$ Here, $a = 2$ and $x = 1.999$.

4. Calculate $f(a)$ and $f'(a)$: $f(2) = 2^4 = 16$ $f'(2) = 4(2^3) = 4 \cdot 8 = 32$

5. Substitute into the approximation: Let $x - a = 1.999 - 2 = -0.001$. Then: $f(1.999) \approx f(2) + f'(2)(x - a)$ $f(1.999) \approx 16 + 32(-0.001)$ $f(1.999) \approx 16 - 0.032$ $f(1.999) \approx 15.968$

Final Answer:

$(1.999)^4 \approx 15.968$

Would you like further details or explanations?

Related Questions:

1. How can linear approximation be applied to functions with square roots?
2. What is the error margin in this approximation for $(1.999)^4$?
3. Can this technique be extended to higher-order approximations?
4. How would this calculation differ if approximating $(2.001)^4$?
5. How does the choice of $a$ affect the accuracy of the linear approximation?

Tip:

When using linear approximation, always choose an $a$ value that is close to $x$ and makes the computation simple!