# Thread: Find an equation for the line tangent to the graph

1. ## Find an equation for the line tangent to the graph

Find an equation for the line tangent to the graph of f(x)=3x(e^x) at the point (a, f(a)) if a=2.

I think you start with taking a derivative. Isn't the derivative just f'(x)=3e^x or do you have to do the product rule? After that, I'm not sure what to do.

2. $\displaystyle f(x) = 3xe^x$

$\displaystyle \frac{df}{dx} = \frac{d}{dx}3xe^x = 3 \frac{d}{dx}xe^x~$ .....

Now use the product rule

3. Ok, so I took the derivative and got f'(x)=3xe^x + 3e^x.

I'm just not sure about how to formulate the equation.

4. The slope of the tangent line of the function $\displaystyle f(x)$ at the point $\displaystyle x=a$ is $\displaystyle f'(a)$.

General formula of a line is $\displaystyle y=mx + C$. $\displaystyle m$ is the slope (which is also f'(a) for the tangent line) and C is a constant. You can find C by plugging $\displaystyle x=a$ and $\displaystyle y=f(a)$ in the formula.

5. Sorry to keep asking questions, but I worked it out and still didn't get it right.

So I plugged 2 into the derivative and got 66.50150489, and then I plugged in a and f(a) and found C to be .3334012213, so I thought my equation would be
y=66.50150489(x+.3334012213), but the math program I'm using is telling me that this is incorrect.

6. Probably because the equation is $\displaystyle y = mx + C$ rather than $\displaystyle y = m (x + C)$ as you used.

7. Even when I used y=66.50150489x + .3334012213, it still says that's incorrect.

8. Originally Posted by A5HLEY
Find an equation for the line tangent to the graph of f(x)=3x(e^x) at the point (a, f(a)) if a=2.

I think you start with taking a derivative. Isn't the derivative just f'(x)=3e^x or do you have to do the product rule? After that, I'm not sure what to do.
Let's start from the beginning:

$\displaystyle f(x)=3x \cdot e^x~\implies~ f'(x)=3x \cdot e^x + 3e^x = 3e^x(x+1)$

If a = 2 then $\displaystyle P(2, 6e^2)$ and the slope of the tangent is:

$\displaystyle m = f'(2)=3 \cdot e^2 \cdot (2+1)=9e^2$

You know a point and the slope of the tangent. Use the point-slope-formula to calculate the equation of the straight line:

$\displaystyle y - 6e^2 = 9e^2(x-2)~\implies~\boxed{y = 9e^2 \cdot x - 12e^2}$

As you may have noticed: The terms I use are a lot shorter than the decimal numbers you use - and they are exact! The decimal numbers are only approximative values.

9. Thank you. You're right... your values make it much easier to understand. Thank you again for your help!