$f'(x)=(3/x)-4; (1,0)$ derivative at a specific point

Question

So I have been trying to figure out how to do this type of problem for quite a while now.

Question States:

The slope $f'(x)$ at each point $(x,y)$ on a curve $y=f(x)$ is given along with a particular point $(a, b)$ on the curve. Use this information to find $f(x)$. $$ f'(x)=\frac{3}{x}-4\quad\text{ at } (1,0) $$ I would appreciate it greatly if someone could explain how to go about this problem or point me in the right direction. My textbook instructions are confusing me greatly on this particular problem.

Answer 1

You know that $f'(x)=\frac 3x - 4$: take the indefinite integral of each side to get

$$f(x)=3\ln x -4x+C$$

where $C$ is an arbitrary constant.

Now you find the value of the constant by substituting the known values of $x$ and $f(x)$ from your given point $(1,0)$: $f(1)=0$, so

$$0=3 \ln 1 -4 \cdot 1 + C$$ $$C=4$$

So your final formula for $f(x)$ is

$$f(x)=3\ln x -4x+4$$

You should check this on a graphing calculator, program, or web page.