Use linear approximation, i.e. the tangent line, to approximate $$ \frac{1}{1.003} $$ as follows: Let $$ f(x)=\frac{1}{x} $$ and find the equation of the tangent line to $$ f(x) $$ at a "nice" point near 1.003. Then use this to approximate $$ \frac{1}{1.003} $$ .

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**Step 1.** Define the function and choose a point for linearization.  
Let $$ f(x)=\frac{1}{x} $$. A "nice" point near 1.003 is $$ a=1 $$.

**Step 2.** Compute $$ f(a) $$ and $$ f'(x) $$.  
We have  
$$
f(1)=\frac{1}{1}=1.
$$  
Differentiate $$ f(x) $$ with respect to $$ x $$:
$$
f'(x)=-\frac{1}{x^2}.
$$  
Then,
$$
f'(1)=-\frac{1}{1^2}=-1.
$$

**Step 3.** Write the equation of the tangent line (linear approximation) at $$ x=a $$.  
The tangent line formula is:
$$
L(x)=f(a)+f'(a)(x-a).
$$  
Substituting the values, we get:
$$
L(x)=1+(-1)(x-1)=1-(x-1)=2-x.
$$

**Step 4.** Use the tangent line to approximate $$ \frac{1}{1.003} $$.  
Since $$ L(x) $$ approximates $$ f(x) $$ near $$ x=1 $$, substitute $$ x=1.003 $$:
$$
L(1.003)=2-1.003=0.997.
$$

Thus, the approximation is:
$$
\frac{1}{1.003} \approx 0.997.
$$
To approximate $$ \frac{1}{1.003} $$ using linear approximation: 1. Let $$ f(x) = \frac{1}{x} $$ and choose $$ a = 1 $$ as the point of approximation. 2. Compute $$ f(1) = 1 $$ and $$ f'(x) = -\frac{1}{x^2} $$, so $$ f'(1) = -1 $$. 3. The tangent line equation is $$ L(x) = 2 - x $$. 4. Substitute $$ x = 1.003 $$ into $$ L(x) $$ to get $$ L(1.003) = 0.997 $$. Therefore, $$ \frac{1}{1.003} \approx 0.997 $$.

Use linear approximation, i.e. the tangent line, to approximate \( \frac{1}{1.003} \) as follows: Let \( f(x)=\frac{1}{x} \) and find the equation of the tangent line to \( f(x) \) at a "nice" point near 1.003. Then use this to approximate \( \frac{1}{1.003} \) .

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