Q. 67 - 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.

$f\text{'}\left(x\right)=\sqrt{3x+1},f\left(0\right)=1$

Step-by-Step Solution

Answer

The function is$f\left(x\right)=\frac{2{\left(3x+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{9}+\frac{7}{9}$

Step 1. Given Information

The given derivative is $f\text{'}\left(x\right)=\sqrt{3x+1},f\left(0\right)=1$

Step 2. Finding antiderivative

$$\begin{gathered} f\left(x\right)=\int \sqrt{3x+1}dx \\ f\left(x\right)=\frac{2}{3}\frac{{\left(3x+1\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3}+C \\ f\left(x\right)=\frac{2{\left(3x+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{9}+C \end{gathered}$$

Step 3. Finding the value of C

Given that $f\left(0\right)=1$

$$\begin{gathered} 1=\frac{2}{9}+C \\ C=1-\frac{2}{9}=\frac{7}{9} \end{gathered}$$

The function become  $f\left(x\right)=\frac{2{\left(3x+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{9}+\frac{7}{9}$