Term 241, MATH 101, Makeup Quiz 2 (1) 1. Let \( f(x) \) be a function such that \( f(1)=1 \) and \( \lim _{x \rightarrow 1} \frac{f(x)}{x} \) the tangent line to the graph of \( f(x) \) at \( x=1 \) is given (a) \( 2 y-x-1=0 \) (b) \( y-2 x+1=0 \) (c) \( 2 y-2 x+3=0 \) (d) \( y-2 x-1=0 \) (e) \( 2 y+x+1=0 \)

To find the equation of the tangent line to the graph of \( f(x) \) at \( x = 1 \), we need to determine the slope of the tangent line at that point. The slope of the tangent line can be found using the limit: \[ \lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = f'(1) \] Given that \( f(1) = 1 \), we can rewrite the limit as: \[ \lim_{x \to 1} \frac{f(x) - 1}{x - 1} = f'(1) \] We also have the information that \( \lim_{x \to 1} \frac{f(x)}{x} \) is provided, but we need to know its value to find the slope. Let's denote this limit as \( L \): \[ L = \lim_{x \to 1} \frac{f(x)}{x} \] Using the fact that \( f(1) = 1 \), we can evaluate this limit: \[ L = \lim_{x \to 1} \frac{f(x)}{x} = \frac{f(1)}{1} = 1 \] This means that: \[ \lim_{x \to 1} \frac{f(x)}{x} = 1 \] Now, we can relate this to the derivative: \[ \lim_{x \to 1} \frac{f(x)}{x} = \lim_{x \to 1} \frac{f(x) - 1 + 1}{x} = \lim_{x \to 1} \left( \frac{f(x) - 1}{x - 1} \cdot \frac{x - 1}{x} + \frac{1}{x} \right) \] As \( x \to 1 \), \( \frac{1}{x} \to 1 \) and \( \frac{f(x) - 1}{x - 1} \to f'(1) \). Therefore, we have: \[ 1 = f'(1) \] Now we have the slope of the tangent line at \( x = 1 \): \[ f'(1) = 1 \] The equation of the tangent line at the point \( (1, f(1)) = (1, 1) \) with slope \( m = 1 \) can be written in point-slope form: \[ y - 1 = 1(x - 1) \] Simplifying this gives: \[ y - 1 = x - 1 \implies y = x \] Now, we need to check which of the given options corresponds to the line \( y = x \): 1. Rearranging \( 2y - x - 1 = 0 \) gives \( y = \frac{1}{2}x + \frac{1}{2} \) (not a match). 2. Rearranging \( y - 2x + 1 = 0 \) gives \( y = 2x - 1 \) (not a match). 3. Rearranging \( 2y - 2x + 3 = 0 \) gives \( y = x - \frac{3}{2} \) (not a match). 4. Rearranging \( y - 2x - 1 = 0 \) gives \( y = 2x - 1 \) (not a match). 5. Rearranging \( 2y + x + 1 = 0 \) gives \( y = -\frac{1}{2}x - \frac{1}{2} \) (not a match). None of the options directly match \( y = x \). However, if we consider the slope of the tangent line, we can see that the slope is \( 1 \), which is consistent with the line \( y = x \). Since none of the options match \( y = x \), it seems there may be an error in the options provided. However, based on the calculations, the tangent line at \( x = 1 \) is indeed \( y = x \).