The Student Room Group

Increasing Functions



Clearly, f(x)f(x) is increasing when f(x)0f'(x) \geq 0.

Here we get that 123x2012-3x^2 \geq 0 hence 4x24 \geq x^2, ie whenever x24x^2 \leq 4.

This is a C1 problem as far as I'm aware. Draw a sketch of x24x^2 - 4 and look for which values of xx the graph is below 0, then their answer shoud be obvious.
Reply 2
Original post by RDKGames
Clearly, f(x)f(x) is increasing when f(x)0f'(x) \geq 0.

Here we get that 123x2012-3x^2 \geq 0 hence 4x24 \geq x^2, ie whenever x24x^2 \leq 4.

This is a C1 problem as far as I'm aware. Draw a sketch of x24x^2 - 4 and look for which values of xx the graph is below 0, then their answer shoud be obvious.

I've just relaised i didn't square root -4

When square rooting on the other side of the inequality, does the directio of the inequality change direction?
Original post by esmeralda123
I've just relaised i didn't square root -4

When square rooting on the other side of the inequality, does the directio of the inequality change direction?


You cannot square root -4... why/how are you even at a point where you need to do that??

Also, I would not advise you to just 'square root both sides of the inequality'
If you do so, then you'd end up with x2x \leq 2. This is not the correct answer, and rather just one half of it. The reason why is simply because if you take x=6x=-6 which satisfies the inequality x2x\leq 2, you do not get that x24x^2 \leq 4 which is the original requirement!
(edited 5 years ago)
Reply 4
Original post by RDKGames
You cannot square root -4... why/how are you even at a point where you need to do that??

Also, I would not advise you to just 'square root both sides of the inequality'
If you do so, then you'd end up with x2x \leq 2. This is not the correct answer, and rather just one half of it. The reason why is simply because if you take x=6x=-6 which satisfied the inequality, you do not get that x24x^2 \leq 4 which is the original requirement!


I did this:

-3x^2 +12>0
-3x^2 > 12
x^2 <-4
The what do I do with the direction of the inequality when I sqaure root and how do I get two solutions of -2 and 2 from this method ?
Original post by esmeralda123
I did this:

-3x^2 +12>0
-3x^2 > 12


Surely -12 on the RHS....
Reply 6
Original post by RDKGames
Surely -12 on the RHS....


Oh yes but still what happens after x^2 < 4
Original post by esmeralda123
Oh yes but still what happens after x^2 < 4


Solve it, however you know how. This shouldn't come as a new thing to you at this point!
Reply 8
Original post by RDKGames
Solve it, however you know how. This shouldn't come as a new thing to you at this point!


But the inequality changes but don't know when

For instance, for question 2 it is x^ > 2 then the inequality remains the same for the positive root x> root 2

For question 9 however, the inequality changes direction for the positive root of 4
This contradicts. I just don't know when to change the direction of the inequality when square rooting?
Original post by esmeralda123
But the inequality changes but don't know when

For instance, for question 2 it is x^ > 2 then the inequality remains the same for the positive root x> root 2

For question 9 however, the inequality changes direction for the positive root of 4
This contradicts. I just don't know when to change the direction of the inequality when square rooting?


If you are getting confused with the signs of the inequality, just do the same thing every time you need to solve one: SKETCH

In Q2, you want to solve x220x^2 - 2 \geq 0. So sketch x22x^2 - 2 and look for the region where this parabola is above the x-axis.

In Q9 you want to solve x240x^2 - 4 \leq 0. So sketch x24x^2 - 4 and look for the region where this parabola is below the x-axis.
Reply 10
Original post by RDKGames
If you are getting confused with the signs of the inequality, just do the same thing every time you need to solve one: SKETCH

In Q2, you want to solve x220x^2 - 2 \geq 0. So sketch x22x^2 - 2 and look for the region where this parabola is above the x-axis.

In Q9 you want to solve x240x^2 - 4 \leq 0. So sketch x24x^2 - 4 and look for the region where this parabola is below the x-axis.


Why did you do less than because I need to show it is an increasing function
Original post by esmeralda123
Why did you do less than because I need to show it is an increasing function


OK, so then work with 4x204-x^2 \geq 0 instead - they're the same thing!

You swap the inequality whenever you multiply/divide by a -ve. So mult by -1 here and get x240x^2 - 4 \leq 0.

Alternatively;

4x24 \geq x^2

    0x24\implies 0 \geq x^2 - 4

    x240\implies x^2 - 4 \leq 0
Reply 12
Original post by RDKGames
OK, so then work with 4x204-x^2 \geq 0 instead - they're the same thing!

You swap the inequality whenever you multiply/divide by a -ve. So mult by -1 here and get x240x^2 - 4 \leq 0.

Alternatively;

4x24 \geq x^2

    0x24\implies 0 \geq x^2 - 4

    x240\implies x^2 - 4 \leq 0


Do you always try and make x^2 positive when it is an increasing function?
Original post by esmeralda123
Do you always try and make x^2 positive when it is an increasing function?


Remove the 'when it is an increasing function' as that's irrelevant in general inequality solving, and the answer is: depends how confident you are.

+ve x2x^2 means a U shaped parabola which is the sketch that you should be familiar with extremely well.

-ve x2x^2 means an upside down parabola which you can also sketch if you want.
Reply 14
f(x)=123x2f'(x) = 12 - 3x^2. Yeah?

Now remember what f(x)f'(x) represents. It's the gradient of f(x)f(x). If the gradient is positive at a point, then the function is increasing at that point.

So when is 123x2012 - 3x^2 \geq 0?
Well, that's a quadratic equation that can be solved by factorising. You should have a basic idea of what it'll look like, it's got a negative x2x^2 coefficient so there's only a finite section of it above the x-axis:
https://www.desmos.com/calculator/leyrwymcy5

Now remember that with this graph, this is a graph of f(x)f'(x) against xx. So f(x)0f'(x) \geq 0 for 2x2-2 \leq x \leq 2. So the gradient is positive and therefore it's an increasing function in the areas between x=2x = -2 and x=2x = 2.

Get it?
Reply 15
Original post by esmeralda123
Do you always try and make x^2 positive when it is an increasing function?


How exactly is x2x^2 going to be negative?
Original post by Sinnoh
How exactly is x2x^2 going to be negative?


Fairly sure they meant whether it's better to have x2x^2 vs x2-x^2 in the inequality, not whether x2<0x^2 < 0.

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