cherylim23's methods question thread :)

[99.95]

how do u get

x>0 and x< -2 fo this question

let f(x) = x^3 + 3x^2 - 1

find: {x:f'(x) > 0}

[m@tty]

That point is not included, for the reason you stated.

But consider the point

and hence it is included.

Similarly, the point is also included. In fact, any point outside of the circle is included, even those to the left and right. Hence there are points involved in the region for all real values of y.

[m@tty]

how do u get

x>0 and x< -2 fo this question

let f(x) = x^3 + 3x^2 - 1

find: {x:f'(x) > 0}

So the intercepts are at and , and it is an upright parabola.

Hence it is greater then zero before the lesser root and above the greater root. ie.

You can sketch to see the result more visually.

[99.95]

thanks m@tty

how do i do this question:

y= X / 1-x

write dy/dx in terms of y

[|ll|lll|]

Oh I think I get it thanks
Basically, they are just asking the y values that are outside the circle for > 9?

However, for option E, since -3 and 3 are not included, they could be counted as outside the circle too?
Or if you don't mind explaining why option E is not right:P

[brightsky]

thanks m@tty

how do i do this question:

y= x / 1-x

write dy/dx in terms of y

....(1)

...(2)

From (1):



Sub that into (2):



Hope this is right..

[/0]

Apart from the sign, the rest seems alright

[99.95]

correct me if i'm wrong:

for linear approximations:

you must always choose the closet value u can square root. u CANNOT choose any number. E.g. square root 120. can't choose 100 and 20. must be 121 - 1

[Mao]

You can choose 100, but your answer won't be accurate.

Generally, choose a pair of values (x,f(x)) that you accurately know to answer to.

[m@tty]

Oh I think I get it thanks
Basically, they are just asking the y values that are outside the circle for > 9?

However, for option E, since -3 and 3 are not included, they could be counted as outside the circle too?
Or if you don't mind explaining why option E is not right:P

Well, what about the point ?? Here the sum of squares is 18, which is greater than nine and hence included in the region. Similarly and and are all included in this region.

Every single point on the plane that lies outside the circle is included in this region. That's because this circle is the exact line for which the sum of squares is 9, or the distance from the origin is three. Therefore every point outside of this circle satisfies the original requirement. See attached diagram.

[99.95]

is this correct:

for approximate change and approximate increase: use hf'(x)

for approximate value: use f(x) + h'f(x)

[TrueTears]

Think about it graphically, the relation denotes all points whose distance from the origin is greater than 3. As . Now, can be recognised as the hypotonuse of a right angled triangle. So the relation includes any point where a line from the origin is greater than 3 units in length.

As seen in the attached image, points of all values of y satisfy this requirement. Therefore the range is .

i like the implicit generator that /0 provided too, so slick

[m@tty]

is this correct:

for approximate change and approximate increase: use hf'(x)

for approximate value: use f(x) + h'f(x)

Yeah. From memory:



So the known value is and the approximate change is .

Think about it graphically, the relation denotes all points whose distance from the origin is greater than 3. As . Now, can be recognised as the hypotonuse of a right angled triangle. So the relation includes any point where a line from the origin is greater than 3 units in length.

As seen in the attached image, points of all values of y satisfy this requirement. Therefore the range is .

i like the implicit generator that /0 provided too, so slick

Indeed, it is very good

[/0]

Huh what implicit generator? o_O

[TrueTears]

implicit grapher man, pr0 pr0 pr0 in the mathematics forum