Find the intervals on which the graph of \( f \) is concave upward, the intervals on which the graph of \( f \) is concave downward, and the inflection points. \( f(x)=\ln \left(x^{2}-4 x+29\right) \)

To determine the concavity of the function \( f(x) = \ln(x^2 - 4x + 29) \), we need to find the second derivative of \( f \) and analyze its sign. ### Step 1: Find the first derivative \( f'(x) \) Using the chain rule, we have: \[ f'(x) = \frac{1}{x^2 - 4x + 29} \cdot (2x - 4) \] This simplifies to: \[ f'(x) = \frac{2x - 4}{x^2 - 4x + 29} \] ### Step 2: Find the second derivative \( f''(x) \) To find the second derivative, we will use the quotient rule on \( f'(x) \): \[ f''(x) = \frac{(x^2 - 4x + 29)(2) - (2x - 4)(2x - 4)}{(x^2 - 4x + 29)^2} \] Calculating the numerator: 1. The first term: \( (x^2 - 4x + 29)(2) = 2x^2 - 8x + 58 \) 2. The second term: \( (2x - 4)(2x - 4) = 4x^2 - 16x + 16 \) Now, combine these: \[ f''(x) = \frac{(2x^2 - 8x + 58) - (4x^2 - 16x + 16)}{(x^2 - 4x + 29)^2} \] Simplifying the numerator: \[ 2x^2 - 8x + 58 - 4x^2 + 16x - 16 = -2x^2 + 8x + 42 \] Thus, we have: \[ f''(x) = \frac{-2x^2 + 8x + 42}{(x^2 - 4x + 29)^2} \] ### Step 3: Find the critical points of \( f''(x) \) To find the intervals of concavity, we need to set the numerator equal to zero: \[ -2x^2 + 8x + 42 = 0 \] Dividing the entire equation by -2 gives: \[ x^2 - 4x - 21 = 0 \] Now, we can factor this quadratic: \[ (x - 7)(x + 3) = 0 \] Thus, the critical points are: \[ x = 7 \quad \text{and} \quad x = -3 \] ### Step 4: Test intervals around the critical points We will test the sign of \( f''(x) \) in the intervals \( (-\infty, -3) \), \( (-3, 7) \), and \( (7, \infty) \). 1. **Interval \( (-\infty, -3) \)**: Choose \( x = -4 \) \[ f''(-4) = \frac{-2(-4)^2 + 8(-4) + 42}{(x^2 - 4x + 29)^2} = \frac{-32 - 32 + 42}{(16 + 16 + 29)^2} = \frac{-22}{(61)^2} < 0 \] (Concave downward) 2. **Interval \( (-3, 7) \)**: Choose \( x = 0 \) \[ f''(0) = \frac{-2(0)^2 + 8(0) + 42}{(0^2 - 4(0) + 29)^2} = \frac{42}{(29)^2} > 0 \] (Concave upward) 3. **Interval \( (7, \infty) \)**: Choose \( x = 8 \) \[ f''(8) = \frac{-2(8)^2 + 8(8) + 42}{(8^2 - 4(8) + 29)^2} = \frac{-128 + 64 + 42}{(64 - 32 + 29)^2} = \frac{-22}{(61)^2} < 0 \] (Concave downward) ### Step 5: Summary of concavity and inflection points - **Concave upward** on the interval \( (-3,