# Find any stationary points of this function.

• May 10th 2012, 11:56 AM
sarahheath1985
Find any stationary points of this function.
Find any stationary points of this function and state whether they are local max or min.
g(x)=(e^(-3x+2sinx)/6)/(3+2cosx) 0<x<2pi included.
• May 10th 2012, 12:51 PM
tom@ballooncalculus
Re: Find any stationary points of this function.
Just in case a picture helps with the quotient rule...

http://www.ballooncalculus.org/draw/diffProd/nine.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside each leg of...

http://www.ballooncalculus.org/asy/prod.png

... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

Ignore the denominator, and the exponential which can't be zero. So you have...

$4 \cos^2 x - 9 + 2 \sin x = 0$

Apply pythagoras.

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• May 10th 2012, 02:01 PM
sarahheath1985
Re: Find any stationary points of this function.
Hi,

Thanks. I can see where I was going wrong now.

How would you apply this to the first derivative test though?

Cheers
• May 10th 2012, 02:50 PM
tom@ballooncalculus
Re: Find any stationary points of this function.
Quote:

Originally Posted by sarahheath1985
How would you apply this to the first derivative test though?

Do you mean the second? Setting the first derivative to zero as we've just done to find the stationary points is the first. (Apply Pythag and solve the quadratic.)

Finding the +/- sign of the second derivative at these points determines whether the curve is, following it left to right, headed ultimately up (because turning out of a minimum) or down (coming out of a maximum). (The 'second derivative test'.)

But determining whether min or max is often (probably here) easier just by testing the values of the original function around the stationary point.