At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

The sign is positive, but decreasing....?

Somewhat, yes, but muddy.

In our specific example, the function is "decreasing" along rows, so \(f_x \) must be negative.

Just pick any one row and forget about everything else.

I'm doing two problems at once.

Sorry for the mess.

thats okay :)

Thanks for understanding.... I'm just so stressed right now.

Decreasing, as we go from left to right.

Yes. so \(f_x(-2,~0)~~\lt ~0 \)
|dw:1435035518143:dw|

Increasing....

good, so \(f_y(-2,~1)\gt 0\)
|dw:1435035643341:dw|

Ohhhhhhh I get the pattern.

is f_xx true, then?

and f_yy

sry i don't seem to remember, il review quick and get back
meantime try other problems maybe..

Okay!

Thanks!

I'm thinking of working the first differences but not entirely sure

|dw:1435037234174:dw|

If it helps, the answer is letter B.

For number 15. I just don't have the solutions.

Answers to the other numbers:
15) B
16) A
17) E
18) A
19) B

Yeah we cannot tell anything about mixed partials i guess
so B makes sense to me

|dw:1435037903299:dw|

we have checked four conditions, so I think option D. Am I right?

namely:
the first and fourth conditions are false;
the second and the third conditions are true

and we can not infer anything on the fifth condition

|dw:1435038498718:dw|

Thanks! How were those numbers selected? Was it through the symmetric difference quotient?

Ohh no, i guess we should use symmetric difference quotient then

Ohhh yeah, I just am confused on how to plug the values in :P.

*am just

Nice! Thanks!

#17 just follows from #16

I see!

Ohhhhhh it's becoming clear to me.... I hope. Haha. I feel like it is.

It really was just a matter of where to look, on the chart....

How do we do number 18? I'm used to solving for it with the equation given.

Do we divide <-2, 3> by \[\sqrt{(-2)^2+(3)^2}\]?

remember, the function value doesn't change when you stay on a specific level curve ?

I think so.

When you're along one level curve, yes....

Also recall the fact that gradient vector is perpendicular to the level curve

Wait, never mind what I said about the root.

Oh yes, it is.

|dw:1435040370688:dw|

|dw:1435040574730:dw|

Ohhhh, yes

Of course. Yes. I did not think about using the given answers.

Wow.

Wait, hmmm what is z?

I can't use the dot product without a z-value....

Ah well. I still get zero, so I guess you can use it without a z-value....

I used 0 for the z-value.

Oh yeah.... -__-

oops

no wait, i got confused

:)

#19 should be easy, just work the tangent plane approximation

And linearization is just: \[L(x, y) = f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)\]?

that should do

or you may use the gradient

normal vector of the required tangent plane is \(\langle f_x,~f_y,~-1 \rangle \)

whichever way you feel easy... upto you

hey still here ?

Hi! I'm back.

This is a simpler method! I like it! Thank You very much, @ganeshie8 ! Hope I do well on my exam!

I'm sure you will do great! good luck!

Thank You!