Q. 6 - Calculus

Question

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

$$\begin{gathered} f\text{'}\left(x\right)=3 \sin (-2x)-5 \cos \left(3x\right), \\ f\left(0\right)=0 \end{gathered}$$

Step-by-Step Solution

Answer

The function f is $f\left(x\right)=\frac{3}{2}\cos (-2x)-\frac{5}{3}\sin \left(3x\right)+c$ and the value of c  is .$\frac{3}{2}$

Step 1. Given Information.

The derivative:
$$\begin{gathered} f\text{'}\left(x\right)=3 \sin (-2x)-5 \cos \left(3x\right), \\ f\left(0\right)=0 \end{gathered}$$

Step 2. Find the function by integration.

Integrating both sides with respect to x, 

$$\begin{gathered} \int f\text{'}\left(x\right).dx=\int 3 \sin (-2x)-5 \cos \left(3x\right)dx \\ f\left(x\right)=\int \sin (-2x)dx-5\int \cos \left(3x\right) dx \\ =3\left(\frac{1}{2}\cos \right(-2x)-5(\frac{1}{3}\sin \left(3x\right)+c \\ =\frac{3}{2}\cos (-2x)-\frac{5}{3}\sin \left(3x\right)+c \end{gathered}$$

where c is a constant.

Step 3. Find the value of c.

Substitute the values in the function, 

$$\begin{gathered} f\left(x\right)=\frac{3}{2}\cos (-2x)-\frac{5}{3}\sin \left(3x\right)+c \\ f\left(0\right)=\frac{3}{2}\cos (-2(0\left)\right)-\frac{5}{3}\sin \left(3\right(0\left)\right)+c \\ 0=\frac{3}{2}\left(1\right)-\frac{5}{3}\left(0\right)+c \\ c=\frac{3}{2} \end{gathered}$$