Question

The turning point of the function $ y = \frac{ax-b}{(x-1)(x-4)} $ at the point $ P(2, -1) $ is

• A. neither a maximum nor a minimum
• B. both maximum and a minimum
• C. a minimum
• D. a maximum

Hint:

To find turning points, always check the first and second derivatives of the function. Use the quotient rule for rational functions, and evaluate the second derivative at the point of interest.

Answer 1

The Correct Option is D

To find the turning point, we need to first calculate the derivative of the function. The function given is: \[ y = \frac{ax-b}{(x-1)(x-4)}. \] We can apply the quotient rule for differentiation: \[ \frac{dy}{dx} = \frac{(x-1)(x-4) \cdot \frac{d}{dx}(ax-b) - (ax-b) \cdot \frac{d}{dx}[(x-1)(x-4)]}{(x-1)^2(x-4)^2}. \] Now, differentiate the numerator and denominator carefully, and substitute \( x = 2 \) to check the nature of the turning point at that point. By analyzing the second derivative or using test points around \( x = 2 \), we can conclude that the turning point at \( P(2, -1) \) is a maximum. 
Thus, the correct answer is (D).