find f such that f(x) = 7x^2 + 5x - 8 and f(0) = 4. f(x) =

Explanation:

Step1: Integrate $f^{\prime}(x)$

We know that if $f^{\prime}(x)=7x^{2}+5x - 8$, then by the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $f(x)=\int(7x^{2}+5x - 8)dx=7\int x^{2}dx+5\int xdx-8\int dx$. So $f(x)=7\times\frac{x^{3}}{3}+5\times\frac{x^{2}}{2}-8x + C=\frac{7}{3}x^{3}+\frac{5}{2}x^{2}-8x + C$.

Step2: Use the initial condition

Given $f(0) = 4$. Substitute $x = 0$ into $f(x)$: $f(0)=\frac{7}{3}(0)^{3}+\frac{5}{2}(0)^{2}-8(0)+C$. Since $f(0) = 4$, then $C = 4$.

Answer:

$\frac{7}{3}x^{3}+\frac{5}{2}x^{2}-8x + 4$