Find the intervals on which the graph of \( f \) is cuncave upwarl \( f(x)=\ln \left(x^{2}-2 x+10\right) \) For what interval(s) of \( x \) is the graph of \( f \) concave upward? A. B. The graph is never concevve upword. For what interval(s) of \( x \) is the grapli of I concave downwars A. \( \square \) (Type your answer in interval notation. Type an exa B. The graph is never concave cownward.

We start with \[ f(x)=\ln\left(x^2-2x+10\right). \] Let \[ g(x)=x^2-2x+10. \] Then \[ f(x)=\ln(g(x)). \] The first derivative is given by \[ f'(x)=\frac{g'(x)}{g(x)}. \] Since \[ g'(x)=2x-2, \] we have \[ f'(x)=\frac{2x-2}{x^2-2x+10}. \] To study the concavity, we compute the second derivative using the quotient rule for derivatives: \[ f''(x)=\frac{g''(x)g(x)-(g'(x))^2}{[g(x)]^2}. \] Since \[ g''(x)=2, \] we substitute: \[ f''(x)=\frac{2\,(x^2-2x+10)-(2x-2)^2}{(x^2-2x+10)^2}. \] Let’s simplify the numerator. First, expand: \[ 2(x^2-2x+10)=2x^2-4x+20, \] and note that \[ (2x-2)^2=4(x-1)^2=4(x^2-2x+1)=4x^2-8x+4. \] Thus the numerator becomes: \[ 2x^2-4x+20- (4x^2-8x+4). \] Simplify term by term: \[ (2x^2-4x^2) + (-4x+8x) + (20-4)= -2x^2+4x+16. \] Factor out \(-2\): \[ -2x^2+4x+16=-2\,(x^2-2x-8). \] The quadratic factors as: \[ x^2-2x-8=(x-4)(x+2). \] Thus, we have: \[ f''(x)=\frac{-2\,(x-4)(x+2)}{(x^2-2x+10)^2}. \] Since the denominator \((x^2-2x+10)^2\) is always positive (because \(x^2-2x+10=(x-1)^2+9>0\) for all \(x\)), the sign of \(f''(x)\) depends solely on the numerator \(-2\,(x-4)(x+2)\). The constant \(-2\) is negative, so the sign of \(f''(x)\) is determined by the sign of \((x-4)(x+2)\). 1. For the graph to be concave upward, we need \(f''(x)>0\). This happens when \[ -2\,(x-4)(x+2) > 0 \quad \Longleftrightarrow \quad (x-4)(x+2) < 0. \] 2. For the graph to be concave downward, we need \(f''(x)<0\), which occurs when \[ (x-4)(x+2) > 0. \] Solve \((x-4)(x+2) < 0\): - This inequality holds when one factor is positive and the other is negative. - When \(x+2>0\) (i.e., \(x>-2\)) and \(x-4<0\) (i.e., \(x<4\)), then \((x-4)(x+2)<0\). Thus, \(f''(x)>0\) for \[ -2 0\): - This happens when both factors are positive or both are negative. - Both positive: \(x-4>0\) and \(x+2>0\) yields \(x>4\). - Both negative: \(x-4<0\) and \(x+2<0\) yields \(x<-2\). Thus, \(f''(x)<0\) for \[ x<-2 \quad \text{or} \quad x>4. \] In summary: - The graph of \(f\) is concave upward on \(\boxed{(-2,4)}\). - The graph of \(f\) is concave downward on \(\boxed{(-\infty,-2)\cup (4,\infty)}\).