question solve the following initial value pr...

# Explanation: ## Step1: Integrate $f''(x)$ to get $f'(x)$ \[ \begin{align*} f'(x)&=\int(48x^{2}-24x + 8)dx\\ &=48\times\frac{x^{3}}{3}-24\times\frac{x^{2}}{2}+8x + C_1\\ &=16x^{3}-12x^{2}+8x + C_1 \end{align*} \] ## Step2: Integrate $f'(x)$ to get $f(x)$ \[ \begin{align*} f(x)&=\int(16x^{3}-12x^{2}+8x + C_1)dx\\ &=16\times\frac{x^{4}}{4}-12\times\frac{x^{3}}{3}+8\times\frac{x^{2}}{2}+C_1x + C_2\\ &=4x^{4}-4x^{3}+4x^{2}+C_1x + C_2 \end{align*} \] ## Step3: Use the initial - condition $f(0)=-6$ Substitute $x = 0$ and $f(0)=-6$ into $f(x)$: \[ \begin{align*} f(0)&=4\times0^{4}-4\times0^{3}+4\times0^{2}+C_1\times0 + C_2\\ -6&=C_2 \end{align*} \] ## Step4: Use the initial - condition $f(-2)=90$ and $C_2=-6$ Substitute $x=-2$, $C_2 = - 6$ into $f(x)$: \[ \begin{align*} f(-2)&=4\times(-2)^{4}-4\times(-2)^{3}+4\times(-2)^{2}+C_1\times(-2)-6\\ 90&=4\times16-4\times(-8)+4\times4-2C_1-6\\ 90&=64 + 32+16-2C_1-6\\ 90&=106-2C_1\\ 2C_1&=106 - 90\\ 2C_1&=16\\ C_1&=8 \end{align*} \] # Answer: $f(x)=4x^{4}-4x^{3}+4x^{2}+8x - 6$