Question

Please help!!!!

Answer 1

@jim_thompson5910

Answer 2

What do you have so far?

Answer 3

i know intervals increasing means as the x-values increases the y-values increase too

Answer 4

@jim_thompson5910

Answer 5

thanks

Answer 6

@jim_thompson5910

Answer 7

hmm trying to think. The fact that they have a y^2 in there is throwing me off. Usually increasing functions are in the form y = f(x)

Answer 8

I guess we could break it up into 2 pieces, but I'm not sure if that will work

Answer 9

I'm thinking there's a typo and it should be y = x^2 - 1 or maybe x^2 = y - 1

Answer 10

ok so what should we do?

Answer 11

wouldn't it just be y= x^2 -1

Answer 12

Assuming there is no typo, we can solve for y to get

y = sqrt(x-1) or y = -sqrt(x-1)

don't forget about the plus/minus when you square root both sides

Answer 13

Graphing the two equations leads to this

https://www.desmos.com/calculator/p3jdxk3zgw

which is a sideways parabola

Answer 14

ok

Answer 15

hopefully you can see that the function is increasing and decreasing on the same interval. Though this isn't a function actually. It fails the vertical line test.

Answer 16

yea i understand

Answer 17

I'm going to ignore the increasing/decreasing stuff. I'm not sure what to make of it

Anyways, onto the domain. What is the smallest x value possible here? How far to the left can we go on the graph?

Answer 18

I would always start out work on a question like this by finding the first and second derivatives, the critical values and the x-values at which we may (or may not) have (a) point (s) of inflection. Have you done these things?

Once you have all these values, we set up intervals on the number line and check the behavior of the function, the 1st derivative and the 2nd derivative on these intervals.

Answer 19

if \[y^2=x-1\]

Answer 20

Use the power and chain rules to obtain dy/dx. In this case, 2y(dy/dx)=1. You must solve for (dy/dx), set it = to 0 and then determine the critical values. And so on.

Answer 21

Maybe it's too late to post this but how about a graphic that is normal size and oriented properly?

Answer 22

what do you mean?

Answer 23

is this better for you?

Answer 24

pfloril234_pop - never mind - I already posted mine.

Answer 25

ok can u help me?

Answer 26

@mathmale so would the first one be (0,2)

Answer 27

Have you already found the first and second derivatives of the given implicit function?
Not clear what you mean by "0,2)."

Answer 28

can you just show me how to do one and then ill do the rest

Answer 29

sorry I'm one of those learners

Answer 30

As before, implicit differentiation of y^2 = x - 1 results in \[2y \frac{ dy }{ dx }=1,\]

Answer 31

which must be solved for dy/dx. Correct result is \[\frac{ dy }{ dx }=\frac{ 1 }{ 2y }\]

Answer 32

You can set this = 0, but there are no solutions. Therefore, your function has no critical values (x-values at which the first derivative = 0).

Answer 33

It may be helpful to you to actually graph this relationship. The graph has the shape of a parabola which passes through (1,0) and opens to the right; it's a horiz. parab. Hint: Graph the top half only, to ease this task. Graph:\[y=+\sqrt{x-1}\]

Answer 34

Then the other half of the graph is this first graph, reflected in the x-axis.

Answer 35

Now, if \[\frac{ dy }{ dx }=\frac{ 1 }{ 2y }, \]

Answer 36

find the second derivative of this with respect to x:\[\frac{ d^2y }{ dx^2 }=\frac{ d }{ dx }\frac{ 1 }{ 2y }\]

Answer 37

Able to follow this discussion? I'd like to hear from you so as to know that you're understanding and are involved.

Answer 38

yes I'm understanding i just get confused using variables like the d and all that...i rather use numbers... like for an example like x-5 where would i put that in the equation u gave me

Answer 39

Caution: the "d's" are not variables.

The symbol for the first derivative (a quantity) is \[\frac{ dy }{ dx }\]

Answer 40

and the operator that produces the derivative of whatever follows it is\[\frac{ d }{ dx}\]

Answer 41

It is critically important that you understand these symbols for derivative operator and derivative.

Answer 42

The operator that produces the 2nd derivative from the 1st derivative is \[\frac{ d }{ dx },\]

Answer 43

same as before.

The symbol for the 2nd derivative of the function y is:\[\frac{ d^2 y }{ d^2 x }\]

Answer 44

supposing that y ou were given the function y=x^2+1, finding the derivative is straightforward: apply the derivative operator d/dx to each of the 3 terms in this equation:\[\frac{ dy }{ dx}=2x\]

Answer 45

However, if we leave this relatively easy example and now must differentiate y^2=x-1, we have to use the Chain Rule, because y is assumed to be a function of the independent variable, x. Make sense?

Answer 46

Here's the process of finding the 2nd derivative of y^2=x-1 from the 1st:

Answer 47

\[\frac{ d }{ dx}\frac{ dy }{ dx }=\frac{ d^2y }{ dx^2 }=\frac{ d }{ dx }\frac{ 1 }{ 2y }\]

Answer 48

Re-writing 1/(2y) as\[(2y)^{-1}\]

Answer 49

and applying the derivative operator, we get \[-1(2y)^{-2}\frac{ d }{ dx}(2y),\]

Answer 50

which can be simplified (the derivative of 2y is\[2\frac{ dy }{ dx }.\]

Answer 51

Homework for you: Please find the simplest possible way in which you can express the second derivative, \[\frac{ d^2y }{ dx^2 }.\]

Answer 52

I need to get off the computer very soon.
Log on to OpenStudy tomorrow and give me a time frame or time frames during which you'd be available to continue this conversation and work out the solution of this problem.

Answer 53

ok thanks for all ur help

Answer 54

Summary: To tackle this problem, you must do the following for beginners:
1. Find the 1st and 2nd derivatives of y with respect to x
2. Find the critical values, if any, by setting the 1st deriv. = to 0 and solving for x (we found none).
3. Find the possible locations of inflection points by setting the 2nd derivative = to 0 and solving for x (if possible).

To Be Continued