1. **State the problem:** Determine for which function the Second Derivative Test cannot conclude that $x=2$ 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 $f'(c)=0$ and $f''(c)<0$, then $f$ has a relative maximum at $c$. If $f''(c)=0$, the test is inconclusive. 3. **Check each function at $x=2$: Find $f'(2)$ and $f''(2)$** **A.** $f(x)=\cos(x-2)$ $f'(x)=-\sin(x-2)$ Evaluate $f'(2)=-\sin(2-2)=-\sin(0)=0$ Find $f''(x)=-\cos(x-2)$ Evaluate $f''(2)=-\cos(0)=-1$ Since $f'(2)=0$ and $f''(2)=-1<0$, Second Derivative Test says relative maximum at $x=2$. **B.** $f(x)=xe^{-1/2}$ (constant factor $e^{-1/2}$) $f'(x)=e^{-1/2} - \frac{1}{2}xe^{-1/2} = e^{-1/2}(1 - \frac{x}{2})$ Evaluate $f'(2)=e^{-1/2}(1 - 1)=0$ Find $f''(x) = -\frac{1}{2}e^{-1/2}$ (constant) Evaluate $f''(2) = -\frac{1}{2}e^{-1/2} < 0$ Second Derivative Test says relative maximum at $x=2$. **C.** $f(x)=x^2 - 4x - 2$ $f'(x)=2x - 4$ Evaluate $f'(2)=2(2)-4=0$ Find $f''(x)=2$ Evaluate $f''(2)=2>0$ Second Derivative Test says relative minimum at $x=2$. **D.** $f(x)=x^3 - 6x^2 + 12x + 1$ $f'(x)=3x^2 - 12x + 12$ Evaluate $f'(2)=3(4) - 24 + 12=12 - 24 + 12=0$ Find $f''(x)=6x - 12$ Evaluate $f''(2)=6(2) - 12=12 - 12=0$ Since $f''(2)=0$, the Second Derivative Test is inconclusive here. 4. **Conclusion:** The Second Derivative Test cannot be used to conclude relative min or max at $x=2$ for function D. **Final answer:** Function D