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$$f(x,y) = \left(y^2 + y -16\right)\sin(x)$$

Find ALL stationary points of $f$ and classify each as local max, min or saddle point.

My working so far is

  • $f_x = \left(y^2 + y -16\right)\cos x$
  • $f_y = \left(2y + 1\right)\sin x$
  • $f_{xx} = -\left(y^2 + y - 16\right)\sin x$
  • $f_{yy}= 2\sin x$
  • $f_{xy}= \left(2y + 1\right)\cos x$

For stationary points I need $f_x=0$ and $f_y=0$

For $\left(2y+1\right)\sin(x) = 0$ need either $y=-\dfrac{1}{2}$ or $x=0$. Now have I made a mistake somewhere because when I put into the other equation to find stationary points when $x = 0$, $y = \dfrac{-1 \pm \sqrt{65}}{2}$ which is fine but when I use $y=-\dfrac{1}{2}$ there is no $x$ value

Thanks in advance!

Harry Peter
  • 8,193
Clarice Sill
  • 75
  • Is there no $x$-value or do all work? – user88595 Jan 15 '14 at 20:09
  • I think I'm looking for 4 stationary points which I thought I could just get from my y=-1/2 or x=0 but I don't know how to do it now I plug into my calculator and error shown up – Clarice Sill Jan 15 '14 at 20:13
  • Sorry I'm new to this site, why has my work been edited to something different then edited back to pretty much the original? which one is correct I'm so confused! :( – Clarice Sill Jan 15 '14 at 20:27

1 Answers1

1

If there are no restrictions on x,

from $f_x=0$ you get that $y=(-1\pm \sqrt{65})/2$ or $x=(2n+1)\frac{\pi}{2}$, where n is any integer, and

from $f_y=0$ you get that $y=-1/2$ or $x=n\pi$, where n is any integer.

Therefore the stationary points are of the form $(n\pi, \frac{-1\pm\sqrt{65}}{2})$ and $((2n+1)\frac{\pi}{2}, -\frac{1}{2}).$

Now you need to test each of these points using the Second Partials Test.

user84413
  • 27,747
  • thank you so much! thought I might have been on the right lines. So is there only 3 stationary points? (well with the possibility of there being infinitive because of n?) – Clarice Sill Jan 15 '14 at 20:59
  • I'm glad this helped, and you're right that there are only 3 stationary points if x is specified to be in $(0,2\pi)$. – user84413 Jan 15 '14 at 21:20
  • x is not specified to be in (0,2pi)

    I got 2 saddle points and one I'm still working on, I'm hoping is correct using fxxfyy-f^2xy

     – Clarice Sill Jan 15 '14 at 22:05
  • If there is no restriction placed on x, then there are infinitely many stationary points (since n can be any integer). You are correct in using $D=f_{xx}f_{yy}-(f_{xy})^2$. – user84413 Jan 15 '14 at 23:36
  • you're so kind, thanks for the help! I just left the answer as the final coordinate could be a min or max depending on whether n is a positive on negative integer. – Clarice Sill Jan 16 '14 at 01:01