1. **Stating the problem:** Analyze the piecewise function defined as: $$f(x) = \begin{cases} \frac{1}{2}x^2 + 3x - 4, & 2 \leq x \leq 4 \\ \ln(5 - x), & 4 < x < 5 \end{cases}$$ with additional geometric elements: a semicircle centered at $\left(\frac{11}{2},0\right)$ and a half branch of a hyperbola with vertex at $(6,0)$ and asymptote at $30^\circ$. We need to find: - Domain and codomain - Points of discontinuity with classification - Stationary points with classification - Points of non-differentiability - Inflection points - Graphs of the first and second derivatives 2. **Domain and codomain:** - For $2 \leq x \leq 4$, $f(x) = \frac{1}{2}x^2 + 3x - 4$ is a polynomial, continuous and differentiable. - For $4 < x < 5$, $f(x) = \ln(5 - x)$ is defined for $x < 5$. - Domain: $[2,5)$ Codomain depends on values: - At $x=2$, $f(2) = \frac{1}{2}(4) + 6 - 4 = 2 + 6 - 4 = 4$ - At $x=4$, $f(4) = \frac{1}{2}(16) + 12 - 4 = 8 + 12 - 4 = 16$ - As $x \to 5^-$, $\ln(5 - x) \to -\infty$ So codomain is $(-\infty, 16]$ approximately. 3. **Points of discontinuity:** Check continuity at $x=4$: - Left limit: $f(4^-) = 16$ - Right limit: $f(4^+) = \ln(5 - 4) = \ln(1) = 0$ - Since $16 \neq 0$, there is a jump discontinuity at $x=4$. 4. **Stationary points:** Calculate derivative for $2 \leq x \leq 4$: $$f'(x) = \frac{d}{dx} \left( \frac{1}{2}x^2 + 3x - 4 \right) = x + 3$$ Set $f'(x) = 0$: $$x + 3 = 0 \Rightarrow x = -3$$ But $x=-3$ is outside domain $[2,4]$, so no stationary points in this interval. For $4 < x < 5$: $$f'(x) = \frac{d}{dx} \ln(5 - x) = -\frac{1}{5 - x}$$ This derivative is never zero, so no stationary points here. 5. **Points of non-differentiability:** At $x=4$, check left and right derivatives: - Left derivative: $f'(4^-) = 4 + 3 = 7$ - Right derivative: $f'(4^+) = -\frac{1}{5 - 4} = -1$ Derivatives differ, so $f$ is not differentiable at $x=4$. 6. **Inflection points:** Second derivative for $2 \leq x \leq 4$: $$f''(x) = \frac{d}{dx} (x + 3) = 1$$ Constant positive, so no inflection points in $[2,4]$. For $4 < x < 5$: $$f''(x) = \frac{d}{dx} \left(-\frac{1}{5 - x}\right) = -\frac{d}{dx} (5 - x)^{-1} = -(-1)(5 - x)^{-2} = -\frac{1}{(5 - x)^2} < 0$$ Second derivative negative, no sign change, so no inflection points here. 7. **Summary:** - Domain: $[2,5)$ - Codomain: $(-\infty,16]$ - Discontinuity: jump at $x=4$ - Stationary points: none in domain - Non-differentiability: at $x=4$ - Inflection points: none 8. **Graphs of derivatives:** - First derivative: - $f'(x) = x + 3$ for $2 \leq x \leq 4$ - $f'(x) = -\frac{1}{5 - x}$ for $4 < x < 5$ - Second derivative: - $f''(x) = 1$ for $2 \leq x \leq 4$ - $f''(x) = -\frac{1}{(5 - x)^2}$ for $4 < x < 5$ These show a linear increasing slope on $[2,4]$ and a negative concavity on $(4,5)$. **Final answer:** The function is continuous and differentiable on $(2,4)$ and $(4,5)$ but has a jump discontinuity and non-differentiability at $x=4$. There are no stationary or inflection points in the domain. The first derivative is piecewise linear and rational negative, and the second derivative is constant positive then negative rational.