question solve the following initial - value ...
question solve the following initial - value problem. f(x)=-x^3 + 1/x^2, f(1)=-21/20 provide your answer below: f(x)=□
Answer
# Explanation:
## Step1: Integrate $f'(x)$
We know that $f(x)=\int f'(x)dx$. Given $f'(x)=-x^{3}+\frac{1}{x^{2}}=-x^{3}+x^{- 2}$. Using the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int(-x^{3}+x^{-2})dx=-\frac{x^{4}}{4}-\frac{x^{-1}}{1}+C=-\frac{x^{4}}{4}-\frac{1}{x}+C$.
## Step2: Use the initial - condition
We are given that $f(1)=-\frac{21}{20}$. Substitute $x = 1$ into $f(x)=-\frac{x^{4}}{4}-\frac{1}{x}+C$. So $f(1)=-\frac{1^{4}}{4}-\frac{1}{1}+C$. Then $-\frac{1}{4}-1 + C=-\frac{21}{20}$. Simplify the left - hand side: $-\frac{1 + 4}{4}+C=-\frac{5}{4}+C$. Now solve for $C$:
\[
\begin{align*}
-\frac{5}{4}+C&=-\frac{21}{20}\\
C&=-\frac{21}{20}+\frac{5}{4}\\
C&=-\frac{21}{20}+\frac{25}{20}\\
C&=\frac{-21 + 25}{20}=\frac{4}{20}=\frac{1}{5}
\end{align*}
\]
## Step3: Write the function $f(x)$
Substitute $C=\frac{1}{5}$ into $f(x)=-\frac{x^{4}}{4}-\frac{1}{x}+C$. We get $f(x)=-\frac{x^{4}}{4}-\frac{1}{x}+\frac{1}{5}$.
# Answer:
$-\frac{x^{4}}{4}-\frac{1}{x}+\frac{1}{5}$