17. Let \( f \) be a differentiable function such that \( f(2) = \frac{5}{2} \) and \( f'(2) = \frac{1}{2} \). If the line tangent to the graph of \( f \) at \( x = 2 \) is used to find an approximation of a zero of \( f \), that approximation is:






Answer is: option1

\( -3 \)

Solution:

The equation of the tangent line at \( x = 2 \) is given by:

\[ y = f(2) + f'(2)(x - 2) \]

Substituting the given values \( f(2) = \frac{5}{2} \) and \( f'(2) = \frac{1}{2} \):

\[ y = \frac{5}{2} + \frac{1}{2} (x - 2) \]

To approximate a root of \( f(x) \), we set \( y = 0 \):

\[ 0 = \frac{5}{2} + \frac{1}{2} (x - 2) \]

1. Subtract \( \frac{5}{2} \) from both sides:

\[ -\frac{5}{2} = \frac{1}{2} (x - 2) \]

2. Multiply both sides by 2 to clear the fraction:

\[ -5 = x - 2 \]

3. Solve for \( x \):

\[ x = -3 \]

The answer is \( -3 \).

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