# slope of a line tangent through a point

I have a problem I am stuck on. its really confusing and I don't know where to start

The slope of the tangent line to the graph of $f(x)$ at each $x\ne 0$ is given by $e^{6x}+{6\over x}$ and knowing that the graph contains the point $(1,{e\over 3})$, find $f(x)$.

How do I start this?

• Do you know how to integrate? You have effectively been told that $f'(x)=e^{6x}+\frac{6}{x}$ and that $f(1)=\frac{e}{3}$. – almagest Jun 15 '16 at 17:53
• i integrated it to (e^6x)/6+6lnx+C so what do i do now? – google Jun 15 '16 at 18:39
• Now you have a curve with that right gradients, but you need to make sure it passes through the given point. So you must pick $C$ so that $f(1)=\frac{e}{3}$. So you want $\frac{e^6}{6}+\ln1+C=\frac{e}{3}$. Hence $C=-\frac{e^6-2e}{6}$. – almagest Jun 15 '16 at 19:29

HINT Note that $$\begin{split} f(x) &= f(1) + \int_{t=1}^{t=x} f'(t) dt \end{split}$$
Integrate the given function (since you obtain the slope by differentiation). You will get a new function +$C$ (constant). This represents a family of curves having that slope at non zero $x$'s. Now since the curve is given to pass through $(1,e/3)$, it must lie on your required curve. Put the point on the curve family and you get your particular curve(ANSWER)