1. **State the problem:** Determine for which function the Second Derivative Test cannot conclude if $x=2$ is a relative minimum or maximum. 2. **Recall the Second Derivative Test:** If $f'(c)=0$ and $f''(c)>0$, then $f$ has a relative minimum at $c$. If $f''(c)<0$, then $f$ has a relative maximum at $c$. If $f''(c)=0$, the test is inconclusive. 3. **Check each function at $x=2$:** **A)** $f(x)=\cos(x-2)$ Calculate $f'(2) = -\sin(2-2) = -\sin(0) = 0$. Calculate $f''(x) = -\cos(x-2)$, so $f''(2) = -\cos(0) = -1 < 0$. Since $f''(2)<0$, Second Derivative Test says relative maximum at $x=2$. **B)** $f(x) = x e^{-x/2}$ Given $f'(x) = e^{-x/2} - \frac{1}{2} x e^{-x/2}$. Calculate $f'(2) = e^{-1} - \frac{1}{2} \cdot 2 \cdot e^{-1} = e^{-1} - e^{-1} = 0$. Calculate $f''(x)$: $$f''(x) = \frac{d}{dx} \left(e^{-x/2} - \frac{1}{2} x e^{-x/2}\right) = -\frac{1}{2} e^{-x/2} - \frac{1}{2} e^{-x/2} + \frac{1}{4} x e^{-x/2} = - e^{-x/2} + \frac{1}{4} x e^{-x/2}$$ Evaluate at $x=2$: $$f''(2) = - e^{-1} + \frac{1}{4} \cdot 2 \cdot e^{-1} = - e^{-1} + \frac{1}{2} e^{-1} = -\frac{1}{2} e^{-1} < 0$$ So $f''(2)<0$, Second Derivative Test says relative maximum. **C)** $f(x) = x^2 - 4x - 2$ Given $f'(x) = 2x - 4$. Calculate $f'(2) = 2(2) - 4 = 0$. Calculate $f''(x) = 2$, so $f''(2) = 2 > 0$. Second Derivative Test says relative minimum at $x=2$. **D)** $f(x) = x^3 - 6x^2 + 12x + 1$ Given $f'(x) = 3x^2 - 12x + 12$. Calculate $f'(2) = 3(4) - 12(2) + 12 = 12 - 24 + 12 = 0$. Calculate $f''(x) = 6x - 12$, so $f''(2) = 6(2) - 12 = 12 - 12 = 0$. Since $f''(2) = 0$, the Second Derivative Test is inconclusive for $x=2$. **Final answer:** The Second Derivative Test cannot be used to conclude for function D.