question solve the following initial value problem. f(x)=48x^2 - 24x + 8, f(0)= - 6, f(-2)=90 provide your…

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:

$f(x)=4x^{4}-4x^{3}+4x^{2}+8x - 6$