91Ó°ÊÓ

The derivative is a key concept used to measure how a function changes as its input changes. It reflects the function's rate of change or slope at a particular point, which is crucial for finding linear approximations. In our original exercise, we need to find the derivative of the function

• Given function: \( f(x) = x^4 \)

Calculating the derivative of \( f(x) = x^4 \) involves using the power rule. The power rule states that

• If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).

For our specific function:- The derivative is \( f'(x) = 4x^3 \).This derivative tells us how the function \( x^4 \) increases as x increases. Knowing this allows us to predict and approximate the function's behavior near any point.

The linearization point is the specific value near which we are approximating our function with a linear function. Choosing a suitable point helps make calculations simpler and the approximation more accurate for nearby values. In our task of estimating \((3.02)^4\):

• The chosen linearization point is \( x = 3 \).

The reason why we choose \( x = 3 \) is because it is close to the value 3.02 that we are interested in, and it makes our calculation straightforward since 3 is an integer.Once we choose this point, we have to compute:\( f'(3) = 4 \times 3^3 \); this gives us 108.This result is essential as it influences the slope of the linear approximation. The precise point of linearization can make a significant difference in the accuracy of approximations for intricate functions.

Function approximation refers to using simpler functions to represent more complex ones over a certain range. With linear approximation, we're using lines (linear functions) to estimate the values of other functions for easier calculations. For our original exercise, this involves constructing a linear function, \( L(x) \), that approximates \( x^4 \) near \( x = 3 \).The formula for local linear approximation is:

• \( L(x) = f(a) + f'(a)(x - a) \)
• Here, \( a \) is the linearization point, \( x \' \) is where we estimate the function, \( f(a) \) is the actual value at \( a \), and \( f'(a) \) is the derivative at \( a \).

For \( x = 3.02 \),

• We calculate \( f(3) = 3^4 = 81 \).
• Then, with our linearization formula: \( L(3.02) = 81 + 108(3.02 - 3) = 81 + 2.16 = 83.16 \).

Thus, the approximated value, using linearization, is \( 83.16 \), which provides a simple estimation instead of performing complex calculations. Function approximation is powerful for quick checks and estimations of values.