If $f^{\prime \prime}(x)=\frac{8}{x^{3}}$ and $f^{\prime}(1)=7$ and $f(1)=3$ then $f(2)=$

Step 1 ：Given that $f''(x) = \frac{8}{x^3}$, we can find $f'(x)$ by integrating $f''(x)$ with respect to $x$.

Step 2 ：So, $f'(x) = \int f''(x) dx = \int \frac{8}{x^3} dx = -\frac{4}{x^2} + C$.

Step 3 ：We know that $f'(1) = 7$, so we can substitute $x = 1$ into the equation $f'(x) = -\frac{4}{x^2} + C$ to find the constant $C$.

Step 4 ：So, $7 = -\frac{4}{1^2} + C$, which simplifies to $7 = -4 + C$.

Step 5 ：Solving for $C$, we get $C = 7 + 4 = 11$.

Step 6 ：So, $f'(x) = -\frac{4}{x^2} + 11$.

Step 7 ：We can find $f(x)$ by integrating $f'(x)$ with respect to $x$.

Step 8 ：So, $f(x) = \int f'(x) dx = \int (-\frac{4}{x^2} + 11) dx = 4x^{-1} + 11x + D$.

Step 9 ：We know that $f(1) = 3$, so we can substitute $x = 1$ into the equation $f(x) = 4x^{-1} + 11x + D$ to find the constant $D$.

Step 10 ：So, $3 = 4(1)^{-1} + 11(1) + D$, which simplifies to $3 = 4 + 11 + D$.

Step 11 ：Solving for $D$, we get $D = 3 - 4 - 11 = -12$.

Step 12 ：So, $f(x) = 4x^{-1} + 11x - 12$.

Step 13 ：Finally, we can find $f(2)$ by substituting $x = 2$ into the equation $f(x) = 4x^{-1} + 11x - 12$.

Step 14 ：So, $f(2) = 4(2)^{-1} + 11(2) - 12 = 2 + 22 - 12 = 12$.

Step 15 ：Therefore, $f(2) = \boxed{12}$.