48. If \( f(x)=x^{2}+1 \) and \( d(x)=\frac{f(x)-f(2)}{x-2} \), then which of the following is true? A. \( d(x)=f(x) \) for all values of \( x \) B. \( d(x)=x-2 \) for all values of \( x \) C. \( d(x)=x+1 \) for all values of \( x \) except 2 D. \( d(2)=f(2) \) E. \( d(x)=x+2 \) for all values of \( x \) except 2

To analyze the function \( d(x) = \frac{f(x) - f(2)}{x - 2} \), we first need to compute \( f(2) \). Given \( f(x) = x^2 + 1 \), we find: \[ f(2) = 2^2 + 1 = 4 + 1 = 5 \] Now, we can express \( d(x) \): \[ d(x) = \frac{f(x) - f(2)}{x - 2} = \frac{f(x) - 5}{x - 2} \] Next, we substitute \( f(x) \): \[ d(x) = \frac{x^2 + 1 - 5}{x - 2} = \frac{x^2 - 4}{x - 2} \] We can factor the numerator: \[ x^2 - 4 = (x - 2)(x + 2) \] Thus, we have: \[ d(x) = \frac{(x - 2)(x + 2)}{x - 2} \] For \( x \neq 2 \), we can simplify this to: \[ d(x) = x + 2 \] However, \( d(x) \) is not defined at \( x = 2 \) because it results in a \( \frac{0}{0} \) form. Therefore, we can conclude that: \[ d(x) = x + 2 \quad \text{for all } x \text{ except } 2 \] Now, let's evaluate the options: A. \( d(x) = f(x) \) for all values of \( x \) - **False**. B. \( d(x) = x - 2 \) for all values of \( x \) - **False**. C. \( d(x) = x + 1 \) for all values of \( x \) except 2 - **False**. D. \( d(2) = f(2) \) - **False** (since \( d(2) \) is not defined). E. \( d(x) = x + 2 \) for all values of \( x \) except 2 - **True**. Thus, the correct answer is: \[ \boxed{E} \]