Q. 68 - Calculus

Question

In Exercises 63–68, find a function that has the given derivative and value. In each case you can find the answer with an educated guess-and-check process while thinking about the chain rule.

$f\text{'}\left(x\right)=\frac{x^2}{\sqrt{x^3+1}},f\left(1\right)=2$

Step-by-Step Solution

Answer

The function is $f\left(x\right)=\frac{2}{3}\sqrt{x^3+1}+\frac{4}{3}$

Step 1. Given Information

The given function is $f\text{'}\left(x\right)=\frac{x^2}{\sqrt{x^3+1}},f\left(1\right)=2$

We need to find antiderivative to get the function.

Step 2. Finding antiderivative

$$\begin{gathered} f\left(x\right)=\int \frac{x^2}{\sqrt{x^3+1}}dx \\ f\left(x\right)=\int \frac{x^2}{\sqrt{t}}\frac{dt}{3x^2}=\frac{1}{3}\int t^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dt \\ f\left(x\right)=\frac{1}{3}\frac{\sqrt{t}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{2}{3}\sqrt{t}+C \\ f\left(x\right)=\frac{2}{3}\left(\sqrt{x^3+1}\right)+C \end{gathered}$$

Step 3. Finding the value of C

Given that $f\left(1\right)=2$

$$\begin{gathered} 2=\frac{2}{3}\sqrt{1}+C \\ C=2-\frac{2}{3}=\frac{4}{3} \end{gathered}$$

The function become $f\left(x\right)=\frac{2}{3}\sqrt{x^3+1}+\frac{4}{3}$