Given that f(x) is a polynomial of degree 4 with f(2)=-1

Given that $f(x)$ is a polynomial of degree 4 with $f(2)=-1,\ f'(2)=0,\ f''(2)=2,\ f'''(2)=-12,\ f^{(4)}(2)=24$, compute $f''(0)$.

Step 1: Reconstruct $f(x)$ about $a=2$

Since $f$ is a degree-4 polynomial, its Taylor expansion about $2$ is exact (the remainder vanishes identically): \[f(x) = \sum_{k=0}^4\frac{f^{(k)}(2)}{k!}(x-2)^k = -1+(x-2)^2-2(x-2)^3+(x-2)^4. \tag{1}\]

Step 2: Differentiate twice

\[ f'(x) = 2(x-2)-6(x-2)^2+4(x-2)^3, \qquad f''(x) = 2-12(x-2)+12(x-2)^2. \tag{2} \]

Step 3: Evaluate at $x=0$

With $x-2=-2$: \[f''(0) = 2-12(-2)+12(-2)^2 = 2+24+48 = 74. \tag{3}\]

Step 4: Verification

\[\text{Direct differentiation of }f(x)=-1+(x-2)^2-2(x-2)^3+(x-2)^4\text{ confirms }f(2)=-1,\ f'(2)=0,\ f''(2)=2,\ f'''(2)=-12,\ f^{(4)}(2)=24\ \checkmark\]