Find stationary point using logs?

[Guest]

I am trying to Find the stationary points of the function, f(x), that lie within the interval :\(\displaystyle \ [ - \pi ,\pi \]\).


\(\displaystyle \L f(x) = (\sin x + \sqrt 3 \cos x)e^{ - x\sqrt 3 } \\)

I found the first st. point by solving \(\displaystyle \L (\sin x + \sqrt 3 \cos x) = 0 \\)

and got .... \(\displaystyle \L x = - \frac{1}{3}\pi \\)

but I am stuck when I try to solve..... \(\displaystyle \L e^{ - x\sqrt 3 } = 0 \\)


I tried using the log:

\(\displaystyle \L - x\sqrt 3 . \ln (e) = 0 \\)

but I can't find a solution to this. What am I doing wrong?

Thank you.

 

[skeeter]

there is no value of x which will make the exponential equal 0 ...

\(\displaystyle \L e^{x\sqrt{3}} > 0\) for all x.

btw ... "stationary points" are where f'(x) = 0, not f(x).

 

[Guest]

Thanks for your reply. Maybe I should re- phrase the question?

I am trying to Find both stationary points of the function, f'(x), that lie within the interval :\(\displaystyle \ [ - \pi ,\pi \]\).

\(\displaystyle \L f'(x) = (\sin x + \sqrt 3 \cos x)-e^{ - x\sqrt 3 } \\)

I understand that this done by equating f'(x) to zero and solving it. I have found one stationery point by solving

\(\displaystyle \L (\sin x + \sqrt 3 \cos x) = 0 \\)

to get .... \(\displaystyle \L x = - \frac{1}{3}\pi \\)

but I don't know how to get the other stationery point.

Can you help?

Thank you.

 

[skeeter]

\(\displaystyle \L \sin{x} + \sqrt{3} \cos{x} = 0\)

\(\displaystyle \L \sin{x} = -\sqrt{3} \cos{x}\)

\(\displaystyle \L \frac{\sin{x}}{\cos{x}} = -\sqrt{3}\)

\(\displaystyle \L \tan{x} = -\sqrt{3}\)

for \(\displaystyle \L -\pi < x < \pi\),

\(\displaystyle \L x = -\frac{\pi}{3}\) , \(\displaystyle \L x = \frac{2\pi}{3}\)