Q. 2 - Calculus

Question

 Find a function f that has the given derivative f'  and value f(c), if possible.  

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

Step-by-Step Solution

Answer

The function is $\frac{2}{3}\ln x+c$ and f(2) is  $2.49970$

Step 1. Given

The given derivative is 

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

Step 2. Integrating the function

$$\begin{gathered} \mathrm{Integrating} \mathrm{both} \mathrm{sides} \\ f\text{'}\left(x\right) = \frac{2}{3x} , f\left(1\right) = 2 \\ ={\int }_{}^{} f\text{'}\left(x\right) =\int \frac{2}{3x} \\ =\frac{2}{3}\ln x+c \\ w\mathrm{here} \mathrm{c} \mathrm{is} \mathrm{constant} \end{gathered}$$

Step 3. Finding the value of f(c)

$$\begin{gathered} f\left(1\right)=2 \\ f\left(x\right)=\frac{2}{3}\ln x+c \\ f\left(1\right)=\frac{2}{3}\ln 1+c \\ 2=0+c \\ c=2 \\ f\left(c\right)=f\left(2\right) \\ f\left(2\right)=\frac{2}{3}\ln 1+c \\ f\left(2\right)=\frac{2}{3}\ln 2+2 \\ f\left(c\right)=2.49970 \end{gathered}$$