find the function $p$ that satisfies the following conditions (assume $x > 0$): $p(x)=\frac{60}{x^{3}}; p(7)…

find the function $p$ that satisfies the following conditions (assume $x > 0$): $p(x)=\frac{60}{x^{3}}; p(7) = 3$. $p(x)=$

find the function $p$ that satisfies the following conditions (assume $x > 0$): $p(x)=\frac{60}{x^{3}}; p(7) = 3$. $p(x)=$

Answer

Explanation:

Step1: Integrate $p'(x)$

We know that if $p'(x)=\frac{60}{x^{2}} = 60x^{- 2}$, then by the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $p(x)=\int60x^{-2}dx=60\frac{x^{-2 + 1}}{-2 + 1}+C=- \frac{60}{x}+C$.

Step2: Use the initial condition $p(7) = 3$

Substitute $x = 7$ and $p(7)=3$ into $p(x)=-\frac{60}{x}+C$. So $3=-\frac{60}{7}+C$. Then solve for $C$: $C=3+\frac{60}{7}=\frac{21 + 60}{7}=\frac{81}{7}$.

Step3: Write the function $p(x)$

Substitute $C=\frac{81}{7}$ back into $p(x)=-\frac{60}{x}+C$, we get $p(x)=-\frac{60}{x}+\frac{81}{7}$.

Answer:

$p(x)=-\frac{60}{x}+\frac{81}{7}$