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I want to find the local max/min and saddle points of $f(x,y) = e^x\cos(y)$.

I started off by finding the following:

\begin{align} f_x &= e^x\cos(y) \\ f_{xx} &= e^x\cos(y) \\ f_y &= -e^x\sin(y) \\ f_{yy} &= -e^x\cos(y) \\ f_{xy} &= -e^x\sin(y) \end{align}

I know I will need the following:

\begin{align} f_x &= 0 \\ f_y &= 0 \end{align}

Now, $\cos(y) = 0$ whenever $y = \frac{\pi}{2}$. Thus, $\frac{\pi}{2}$ would make $f_x = 0$.

Am I proceeding in the right direction?

idkmybffjill
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1 Answers1

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Here is a plot of $f(x,y)$ (maybe it's useful?):

enter image description here

Edit

Since there exists no $x$ that satisfies both $f_x=0$ and $f_y=0$, there are no critical points nor saddle points.

rae306
  • 9,742
  • Okay. I also know that $\sin(y) = 0$ when $y = 0 + 2\pi n$, which tells me when $f_y = 0$. I'm a bit lost where to go from here. – idkmybffjill Sep 16 '14 at 19:23
  • Well, I think my problem is that I have the $y$-coordinates of two different points, right? Namely, $(x_1, \frac{\pi}{2} + 2\pi n)$ and $(x_2, 0 + 2\pi n)$. I need to find what $x_1$ and $x_2$ are before I proceed to the second derivative test? – idkmybffjill Sep 16 '14 at 19:26