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. \( f(x)=\ln \left(x^{2}-2 x+10\right) \) For what interval(s) of \( x \) is the graph of \( f \) concave upward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. B. The graph is never concave upward. For what interval(s) of \( x \) is the graph of \( f \) concave downward? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \square \) (Type your answer in interval notation. Type an exact answer. Use a comma to separate answers as needed.) B. The graph is never concave downward. Determine the \( x \) coordinates of any inflection points of the graph of \( f(x) \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( x=\square \) (Type an exact answer. Use a comma to separate answers as needed.)

We start with \[ f(x)=\ln\left(x^2-2x+10\right) \] **Step 1. Compute the first derivative.** Let \[ u(x)=x^2-2x+10 \quad \text{so that} \quad f(x)=\ln(u(x)). \] Then \[ u'(x)=2x-2, \] and using the chain rule: \[ f'(x)=\frac{u'(x)}{u(x)}=\frac{2x-2}{x^2-2x+10}. \] **Step 2. Compute the second derivative.** Differentiate \( f'(x)=\frac{2x-2}{x^2-2x+10} \) using the quotient rule: \[ f''(x)=\frac{(2)(x^2-2x+10)-(2x-2)(2x-2)}{(x^2-2x+10)^2}. \] Simplify the numerator: \[ \begin{aligned} 2(x^2-2x+10) &= 2x^2-4x+20, \\ (2x-2)^2 &= 4x^2-8x+4, \\ \text{Numerator} &= \,\, 2x^2-4x+20 - (4x^2-8x+4) \\ &= -2x^2+4x+16. \end{aligned} \] Thus, \[ f''(x)=\frac{-2x^2+4x+16}{(x^2-2x+10)^2}. \] **Step 3. Factor the numerator.** Factor out \(-2\): \[ -2x^2+4x+16=-2(x^2-2x-8). \] The quadratic \(x^2-2x-8\) factors further. Its discriminant is \[ \Delta = (-2)^2-4(1)(-8)=4+32=36, \] and its roots are \[ x=\frac{2\pm\sqrt{36}}{2}=\frac{2\pm6}{2}, \] which gives \(x=4\) and \(x=-2\). Therefore: \[ x^2-2x-8=(x-4)(x+2), \] and \[ f''(x)=-\frac{2(x-4)(x+2)}{(x^2-2x+10)^2}. \] Since the denominator \((x^2-2x+10)^2\) is always positive for all \(x\) (because \(x^2-2x+10\) has a negative discriminant), the sign of \(f''(x)\) depends on the numerator \(-2(x-4)(x+2)\). **Step 4. Determine the intervals of concavity.** 1. \(f\) is concave upward when \(f''(x)>0\). Since \(-2\) is negative, we require \[ (x-4)(x+2) < 0. \] The product \((x-4)(x+2)\) is negative when \(x\) is between the roots: \[ -2 < x < 4. \] 2. \(f\) is concave downward when \(f''(x)<0\). This happens when \[ (x-4)(x+2) > 0, \] which occurs for \[ x < -2 \quad \text{or} \quad x > 4. \] **Step 5. Identify the inflection points.** Inflection points occur where \(f''(x)=0\) and the concavity changes. Since \[ f''(x)=-\frac{2(x-4)(x+2)}{(x^2-2x+10)^2}, \] we set the numerator equal to zero: \[ -2(x-4)(x+2)=0 \quad \Longrightarrow \quad x-4=0 \quad \text{or} \quad x+2=0. \] Thus, the possible inflection points are at \[ x=4 \quad \text{and} \quad x=-2. \] Checking the sign of \(f''(x)\) on either side of these points confirms the change in concavity, so both are indeed inflection points. **Answers:** For what interval(s) of \(x\) is the graph of \(f\) concave upward? - The graph is concave upward on \(\boxed{(-2,4)}\). For what interval(s) of \(x\) is the graph of \(f\) concave downward? - The graph is concave downward on \(\boxed{(-\infty, -2) \cup (4, \infty)}\). Determine the \(x\)-coordinates of any inflection points: - Inflection points occur at \(\boxed{x=-2,\, 4}\).