Questions: Given the function f(x) = 5/x + 5/x^2 and the point P(4, 25/16), which lies on the graph of f(x), determine the equation of the tangent line to f(x) at P. The equation of the tangent line to f(x) at P is y= .

Solution Steps

To determine the equation of the tangent line to the function \( f(x) = \frac{5}{x} + \frac{5}{x^2} \) at the point \( P(4, \frac{25}{16}) \), we need to follow these steps:

1. Compute the derivative \( f'(x) \) to find the slope of the tangent line at any point \( x \).
2. Evaluate the derivative at \( x = 4 \) to get the slope of the tangent line at \( P \).
3. Use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point \( P \).

The function is given by

\[ f(x) = \frac{5}{x} + \frac{5}{x^2}. \]

To find the slope of the tangent line, we first compute the derivative \( f'(x) \):

\[ f'(x) = -\frac{5}{x^2} - \frac{10}{x^3}. \]

Next, we evaluate the derivative at the point \( x = 4 \):

\[ f'(4) = -\frac{5}{4^2} - \frac{10}{4^3} = -\frac{5}{16} - \frac{10}{64} = -\frac{5}{16} - \frac{5}{32} = -\frac{15}{32}. \]

Thus, the slope of the tangent line at point \( P(4, \frac{25}{16}) \) is

\[ m = -\frac{15}{32}. \]

Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), we substitute \( m = -\frac{15}{32} \), \( x_1 = 4 \), and \( y_1 = \frac{25}{16} \):

\[ y - \frac{25}{16} = -\frac{15}{32}(x - 4). \]

Rearranging this gives:

\[ y = -\frac{15}{32}x + \frac{15 \cdot 4}{32} + \frac{25}{16}. \]

Calculating \( \frac{15 \cdot 4}{32} = \frac{60}{32} = \frac{15}{8} \) and converting \( \frac{25}{16} \) to eighths gives \( \frac{25}{16} = \frac{50}{32} \). Thus, we have:

\[ y = -\frac{15}{32}x + \frac{15}{8} + \frac{50}{32} = -\frac{15}{32}x + \frac{3.4375}{1}. \]

Final Answer