Questions: Question 15 Identify the function's local and absolute extreme values, if any, saying where they occur. h(x) = (x-1)/(x^2+3x+5) no local extrema local minimum at x=-3; local maximum at x=4 local minimum at x=-2 ; no local maxima local minimum at x=-2; local maximum at x=4

Solution Steps

To find the local and absolute extreme values of the function \( h(x) = \frac{x-1}{x^2+3x+5} \), we need to:

1. Compute the derivative \( h'(x) \) to find critical points by setting \( h'(x) = 0 \).
2. Use the second derivative test or the first derivative test to determine the nature of each critical point (local minimum, local maximum, or neither).
3. Evaluate the function at critical points and endpoints (if any) to find absolute extrema.

We start with the function \( h(x) = \frac{x-1}{x^2 + 3x + 5} \). The first derivative is calculated as follows: \[ h'(x) = \frac{(-2x - 3)(x - 1) + (1)(x^2 + 3x + 5)}{(x^2 + 3x + 5)^2} \]

Setting the first derivative \( h'(x) = 0 \) allows us to find the critical points. Solving this gives us: \[ \text{Critical Points: } x = -2, \, 4 \]

Next, we compute the second derivative \( h''(x) \) to classify the critical points: \[ h''(x) = \frac{(-4x - 6)(-2x - 3)(x - 1) + 2(-2x - 3)(x^2 + 3x + 5) - 2(x - 1)(x^2 + 3x + 5)}{(x^2 + 3x + 5)^3} \] Evaluating \( h''(-2) \) and \( h''(4) \):

• For \( x = -2 \): \( h''(-2) > 0 \) indicates a local minimum.
• For \( x = 4 \): \( h''(4) < 0 \) indicates a local maximum.

Final Answer