I don't get this question. (2 Viewers)

[Dragonmaster262]

Solve simultaneously:

 

[Trebla]

Solve simultaneously:

Hint:
Solve | 2x + 3 | ≥ 2 and | 2x + 3 | ≤ 11 separately, and find the intersection of the two sets of solutions found in each case.

 

[Drongoski]

As suggested by Trebla:




I use a number line to mark off the various boundary points, with -3/2 the point of reference.

 

[Dragonmaster262]

Hint:
Solve | 2x + 3 | ≥ 2 and | 2x + 3 | ≤ 11 separately, and find the intersection of the two sets of solutions found in each case.

Does that count as solving simultaneously?

 

[Dragonmaster262]

A few questions:

Q) How do you know where the graph of a function is positive/negative or zero? When a function is negative is it said to be decreasing and vice versa?

Q) What does it mean, when it you say that a function is bounded?

Q) How do you find the horizontal asymptoe of a function? If it has any that is.

Worded answers would be suffice. There's no need to be using Latex.

 

[GUSSSSSSSSSSSSS]

1) To find where it is positive/negative/zero u sub in x values, and if the resulting f(x) value will be positive/negativ/zero. NO it DOES NOT mean it's decreasing..HOWEVER if f'(x) < 0 (the first derivative is less then zero) the the function IS decreasing as the gradient is negative

2) When a function is bounded it literally means that it has boundaries, it does not go to infinity

3) horizontal asymptotes can be found by either dividing the function (polynomial divisions) and taking the limit as x --> infinity, or by simply by looking at the leading term and its coefficients...

 

[GUSSSSSSSSSSSSS]

i hope that helped....my (3) answer was sort of dodgy sory =S

 

[undalay]

Q) How do you know where the graph of a function is positive/negative or zero? When a function is negative is it said to be decreasing and vice versa?

Q) What does it mean, when it you say that a function is bounded?

Q) How do you find the horizontal asymptoe of a function? If it has any that is.

Worded answers would be suffice. There's no need to be using Latex.

1. Generally by looking at the equation and finding roots, you can tell where the graph is positive/negative (roots are zero). You can test points around the root if you're not sure.
When a function is negative it is not decreasing. Decreasing means the gradient is negative (or if you like when the derivative function is negative)

2. Function is bounded basically refers to definite integrals (I'm assuming) Basically when you have to sub in numbers after you've actually integrated the function.

3. The best way would be to use limits to find horizontal asymptotes. Using limits when x->inf and x-> - inf.

 

[Dragonmaster262]

1) To find where it is positive/negative/zero u sub in x values, and if the resulting f(x) value will be positive/negativ/zero. NO it DOES NOT mean it's decreasing..HOWEVER if f'(x) < 0 (the first derivative is less then zero) the the function IS decreasing as the gradient is negative

2) When a function is bounded it literally means that it has boundaries, it does not go to infinity

3) horizontal asymptotes can be found by either dividing the function (polynomial divisions) and taking the limit as x --> infinity, or by simply by looking at the leading term and its coefficients...

So how do you tell if a function is postive, negative or zero by just looking at the function graph?

Is it still bounded if it has discontinuities in between but still continues along the x and y axes?

4) If a function is at a discontinuity on the x axis, is it said to be constant?

 

[GUSSSSSSSSSSSSS]

So how do you tell if a function is postive, negative or zero by just looking at the function graph?

Is it still bounded if it has discontinuities in between but still continues along the x and y axes?

4) If a function is at a discontinuity on the x axis, is it said to be constant?

1) When it is ABOVE the x axis it is positive, when it is BELOW the x axis it is negative, when it TOUCHES the x axis it is zero

NO the function is UNBOUNDED if it continues on to infinity

a constant funtion is a vertical or horizontal line ie: x = 4, or y = 6
so NO it is not said to be constant, a function with a discontinuity would be known as a discontinuous function

 

[Timothy.Siu]

So how do you tell if a function is postive, negative or zero by just looking at the function graph?

Is it still bounded if it has discontinuities in between but still continues along the x and y axes?

4) If a function is at a discontinuity on the x axis, is it said to be constant?

dont get any of those questions

 

[Dragonmaster262]

dont get any of those questions

This is why numeracy alone isn't enough to bypass maths.

Thanks GUESSSSSS.

 

[undalay]

This is why numeracy alone isn't enough to bypass maths.

lol

 

[lyounamu]

This is why numeracy alone isn't enough to bypass maths.

Thanks GUESSSSSS.

uh what?

You don't know how smart tim is, he gotta be one of the top 09 maths wizards here along with lolokay, jetblack and gurmies.

 

[Dragonmaster262]

Wait, there's one question I forgot to ask. How do you tell if a function is increasing/decreasing by just looking at its graph?

 

[Dragonmaster262]

uh what?

You don't know how smart tim is, he gotta be one of the top 09 maths wizards here along with lolokay, jetblack and gurmies.

I never said he was dumb. I was just cracking a joke. Sorry, if I offended you Tim.

 

[lyounamu]

Wait, there's one question I forgot to ask. How do you tell if a function is increasing/decreasing by just looking at its graph?

increasing -> the graph is heading right and up

decreasing -> the graph is heading right and down

 

[Dragonmaster262]

increasing -> the graph is heading right and up

decreasing -> the graph is heading right and down

Thanks Namu. Finally, this thread is over.

 

[Drongoski]

Bounded Functions

A function f(x) is said to be bounded over its domain if there is a positive number k such that:

I f(x) I <= k (how the heck do u get the abs value bars?)

e.g. f(x) = 3 + sin x

Then: I f(x) I <= 4
in fact I f(x) I < 5
I f(x) I < 5000000

So here we have indicated that f(x) is bounded by 4, 5, 5000000 and 23, 307, .. ..
But of all the bounds in this case, 4 is the least upper bound or the LUB

A function is 'bounded' therefore simply means its value is limited (bounded) . . . i.e. no less than a number 'm' and no greater than a number 'n' say.

 

[drowning_pocoyo]

2 < l2x + 3l
2^2 < (2x + 3)^2
4 < 4x^2 + 12x + 9
0 < 4x^2 + 12x + 5
0 < (2x + 5)(2x + 1)
x < -5/2 and x > -1/2

l2x + 3l < 11
(2x + 3)^2 < 11^2
4x^2 + 12x + 9 < 121
4x^2 + 12x - 112 < 0
4 (x^2 + 3x - 24) < 0
4 (x + 5)(x - 3) < 0
-5 < x < 3

So when they intersect those are your solutions:
-5 < x < -5/2 AND -1/2 < x < 3 hope it helped : )