1. **Problem Statement:** Find the second derivative $f''(x)$, the ratio $\frac{f'(x)}{f(x)}$, and the expression $f''(x) - 2$ for the function $f(x)$ defined by the triangle vertices $(0,0)$, $(3,3)$, and $(6,0)$. 2. **Step 1: Define the function $f(x)$** The triangle vertices suggest a piecewise linear function: - From $x=0$ to $x=3$, the line passes through $(0,0)$ and $(3,3)$, so slope $m_1 = \frac{3-0}{3-0} = 1$. - From $x=3$ to $x=6$, the line passes through $(3,3)$ and $(6,0)$, so slope $m_2 = \frac{0-3}{6-3} = -1$. Thus, $$ f(x) = \begin{cases} x & 0 \leq x \leq 3 \\ - x + 6 & 3 < x \leq 6 \end{cases} $$ 3. **Step 2: Find $f'(x)$** Since $f(x)$ is piecewise linear, $$ f'(x) = \begin{cases} 1 & 0 < x < 3 \\ -1 & 3 < x < 6 \end{cases} $$ Note: $f'(x)$ is not defined at $x=3$ due to the corner. 4. **Step 3: Find $f''(x)$** The derivative of a constant slope is zero everywhere except at the corner where the slope changes abruptly. Thus, $$ f''(x) = 0 \quad \text{for} \quad x \neq 3 $$ At $x=3$, $f''(x)$ is undefined (a Dirac delta in distribution sense), but for typical calculus, $$ f''(x) = 0 \quad \text{almost everywhere} $$ 5. **Step 4: Compute $\frac{f'(x)}{f(x)}$** For $0 < x < 3$: $$ \frac{f'(x)}{f(x)} = \frac{1}{x} $$ For $3 < x < 6$: $$ \frac{f'(x)}{f(x)} = \frac{-1}{-x + 6} = \frac{1}{x - 6} $$ Note: At $x=0$ or $x=3$ or $x=6$, the expression is undefined due to division by zero or derivative discontinuity. 6. **Step 5: Compute $f''(x) - 2$** Since $f''(x) = 0$ almost everywhere, $$ f''(x) - 2 = 0 - 2 = -2 \quad \text{for} \quad x \neq 3 $$ **Final answers:** - $f''(x) = 0$ for $x \neq 3$ - $\frac{f'(x)}{f(x)} = \begin{cases} \frac{1}{x} & 0 < x < 3 \\ \frac{1}{x - 6} & 3 < x < 6 \end{cases}$ - $f''(x) - 2 = -2$ for $x \neq 3$ These results reflect the piecewise linear nature of $f(x)$ and its derivatives.