Answer is: option3

1.7

Solution:

We are given:

\( f(x) = x^2 - 4x + 5 \).

Differentiate:

\( f'(x) = 2x - 4 \).

At \( x = 1 \):

\( f(1) = (1)^2 - 4(1) + 5 = 1 - 4 + 5 = 2 \).

\( f'(1) = 2(1) - 4 = 2 - 4 = -2 \).

Using the tangent line equation:

\( y \approx f(1) + f'(1)(x - 1) \).

\( y \approx 2 + (-2)(x - 1) \).

\( y \approx 2 - 2(x - 1) \).

\( y \approx 2 - 2x + 2 \).

\( y \approx 4 - 2x \).

We need to find the largest \( x \) such that the error is less than 0.5:

\( |f(x) - (4 - 2x)| < 0.5 \).

Compute \( f(x) \):

\( f(x) = x^2 - 4x + 5 \).

Substituting \( f(x) - (4 - 2x) \):

\( |(x^2 - 4x + 5) - (4 - 2x)| < 0.5 \).

\( |x^2 - 4x + 5 - 4 + 2x| < 0.5 \).

\( |x^2 - 2x + 1| < 0.5 \).

\( |(x - 1)^2| < 0.5 \).

\( (x - 1)^2 < 0.5 \).

Taking the square root:

\( |x - 1| < \sqrt{0.5} \).

Approximating \( \sqrt{0.5} \approx 0.707 \), we get:

\( -0.707 < x - 1 < 0.707 \).

\( 1 - 0.707 < x < 1 + 0.707 \).

\( 0.293 < x < 1.707 \).

The greatest \( x \) within this range is 1.7

so the correct answer is: 1.7