Answer is: option2

\( \frac{79}{160} \)

Solution:

We will approximate the function \( y = \frac{1}{\sqrt{x}} \) at \( x = 4.1 \) using the tangent line at \( x = 4 \).

Given:

\( f(x) = \frac{1}{\sqrt{x}} \)

At \( x = 4 \):

\( f(4) = \frac{1}{\sqrt{4}} = \frac{1}{2} \)

Rewrite in Exponent Form:

\( f(x) = x^{-\frac{1}{2}} \)

Differentiate using Power Rule:

\( f'(x) = -\frac{1}{2} x^{-\frac{3}{2}} \)

\( f'(x) = -\frac{1}{2} \times \frac{1}{x^{\frac{3}{2}}} \)

\( f'(x) = -\frac{1}{2x^{\frac{3}{2}}} \)

At \( x = 4 \):

\( f'(4) = -\frac{1}{2(4^{\frac{3}{2}})} \)

Since \( 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8 \), we get:

\( f'(4) = -\frac{1}{2 \times 8} = -\frac{1}{16} \)

Equation of Tangent Line at \( x = 4 \):

\( y \approx f(4) + f'(4)(x - 4) \)

Substituting values:

\( y \approx \frac{1}{2} + \left( -\frac{1}{16} \right)(x - 4) \)

For \( x = 4.1 \):

\( y \approx \frac{1}{2} + \left( -\frac{1}{16} \right)(4.1 - 4) \)

\( y \approx \frac{1}{2} - \frac{1}{16} \times 0.1 \)

\( y \approx \frac{1}{2} - \frac{0.1}{16} \)

Simplifying:

Since \( \frac{0.1}{16} = \frac{1}{160} \), we get:

\( y \approx \frac{1}{2} - \frac{1}{160} \)

\( y \approx \frac{80}{160} - \frac{1}{160} \)

\( y \approx \frac{79}{160} \)

The correct answer is \( \frac{79}{160} \).