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. $f(x) = 3xe^x$

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

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 $f(x)$ at the point $x=a$ is $f'(a)$.

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

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

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

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

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

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

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!