question solve the following initial value pr...

# Explanation: ## Step1: Integrate $f''(x)$ to get $f'(x)$ Integrate $f''(x)=- 36x^{2}-18x - 4$ with respect to $x$. Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $f'(x)=\int(-36x^{2}-18x - 4)dx=-36\times\frac{x^{3}}{3}-18\times\frac{x^{2}}{2}-4x + C=-12x^{3}-9x^{2}-4x + C$. ## Step2: Use the condition $f'(-1) = 13$ to find $C$ Substitute $x=-1$ and $f'(-1)=13$ into $f'(x)=-12x^{3}-9x^{2}-4x + C$. Then $13=-12\times(-1)^{3}-9\times(-1)^{2}-4\times(-1)+C$. Simplify the right - hand side: $13 = 12-9 + 4+C$, $13=7 + C$, so $C = 6$. Thus, $f'(x)=-12x^{3}-9x^{2}-4x + 6$. ## Step3: Integrate $f'(x)$ to get $f(x)$ Integrate $f'(x)=-12x^{3}-9x^{2}-4x + 6$ with respect to $x$. Using the power - rule of integration, $f(x)=\int(-12x^{3}-9x^{2}-4x + 6)dx=-12\times\frac{x^{4}}{4}-9\times\frac{x^{3}}{3}-4\times\frac{x^{2}}{2}+6x+D=-3x^{4}-3x^{3}-2x^{2}+6x + D$. ## Step4: Use the condition $f(-1)=-35$ to find $D$ Substitute $x = - 1$ and $f(-1)=-35$ into $f(x)=-3x^{4}-3x^{3}-2x^{2}+6x + D$. Then $-35=-3\times(-1)^{4}-3\times(-1)^{3}-2\times(-1)^{2}+6\times(-1)+D$. Simplify the right - hand side: $-35=-3 + 3-2-6+D$, $-35=-8 + D$, so $D=-27$. # Answer: $f(x)=-3x^{4}-3x^{3}-2x^{2}+6x - 27$