Questions: Find the open intervals where the function is concave upward or concave downward. Find any inflection points. f(x) = 5/(x-1) Where is the function concave upward and where is it concave downward? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function is concave upward on the interval(s) . The function is concave downward on the interval(s) 1. (Type your answers in interval notation. Use integers or fractions for any numbers in the expressions. Use a comma to separate answers as needed.) B. The function is concave upward on the interval(s) . The function is never concave downward. (Type your answer in interval notation. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) C. The function is concave downward on the interval(s) . The function is never concave upward. (Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) D. The function is neither concave downward nor concave upward.

Solution Steps

To determine where the function is concave upward or downward, we need to find the second derivative of the function \( f(x) = \frac{5}{x-1} \). The concavity of the function is determined by the sign of the second derivative: if it is positive, the function is concave upward; if it is negative, the function is concave downward. We also need to check for any points where the second derivative is zero or undefined, as these could be inflection points.

To determine the concavity of the function, we first need to find the first derivative of the function \( f(x) = \frac{5}{x-1} \).

Using the quotient rule, where if \( f(x) = \frac{u}{v} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \), we have:

• \( u = 5 \) and \( v = x-1 \)
• \( u' = 0 \) and \( v' = 1 \)

Thus, the first derivative is: \[ f'(x) = \frac{0 \cdot (x-1) - 5 \cdot 1}{(x-1)^2} = \frac{-5}{(x-1)^2} \]

Next, we find the second derivative to determine the concavity: \[ f'(x) = \frac{-5}{(x-1)^2} \]

Using the chain rule and power rule, the second derivative is: \[ f''(x) = \frac{d}{dx}\left(-5(x-1)^{-2}\right) = -5 \cdot (-2)(x-1)^{-3} \cdot 1 = \frac{10}{(x-1)^3} \]

The concavity of the function is determined by the sign of the second derivative \( f''(x) \).

• \( f''(x) = \frac{10}{(x-1)^3} \)

1. Concave Upward: \( f''(x) > 0 \) \[ \frac{10}{(x-1)^3} > 0 \implies (x-1)^3 > 0 \implies x-1 > 0 \implies x > 1 \]

2. Concave Downward: \( f''(x) < 0 \) \[ \frac{10}{(x-1)^3} < 0 \implies (x-1)^3 < 0 \implies x-1 < 0 \implies x < 1 \]

An inflection point occurs where the concavity changes, which is where \( f''(x) = 0 \) or is undefined. Here, \( f''(x) \) is undefined at \( x = 1 \), but it does not change sign at this point, so there is no inflection point.

Final Answer