Inequalities graph [Solved!]

X

hi. i met a problem while studying. could u help me solve?

sketch a region A defined by y > x2-2, y < x and x>1. thanks i wait to hear from u

Relevant page

<a href="/inequalities/inequality.php">Inequalities</a>

What I've done so far

I couldn't graph any of it and don't know what to do with x>1

X

Hi Christian

OK, first step to get you started - what does `y = x^2 - 2` look like? 

[Hint: <a href="/quadratic-equations/4-graph-quadratic-function.php">4. The Graph of the Quadratic Function</a>]

Do a sketch of that curve first then think about whether the region `y &gt; x^2 - 2` is above the curve or below (hint, substitute some `(x,y)` values from above and below the curve to see which ones work.

You need to do the same for the line `y = x`.

Finally, if `x &gt; 1`, it just means the x value for every point must be greater than one.

Your final answer is where all 3 regions intersect.

All the best with it. Feel free to get back to me if you are stuck.

X

thanks but can u just sketch so i can have a look. I've tried but got stuck already.hope to hear from u

X

Hi again Christian

If I just give you the answer, I'm worried you won't learn much.

I'm happy to walk you through it. What did your curve `y = x^2 - 2` look like? We'll go step by step.

X

Is it a U shape curve going through `-2` on the y-axis and `-sqrt(2)` and `sqrt(2)` on the x-axis?

X

Hi again

Ah - now we are getting somewhere. It sounds like you have the correct parabola in the correct position.

The > and < are no big deal - they are a simple concept that mathematicians have managed to make look difficult. ^^


In an earlier post I said:

<blockquote>"Think about whether the region y &gt; x^2 - 2 is above the curve or below (hint, substitute some (x,y) values from above and below the curve to see which ones work."</blockquote>

Another hint is that on one side it should work and on the other it should not.

For example, the point (1,5) is in the region above the curve, right? (I just picked a convenient point.) Let's test that point (`x = 1` and `y = 5`) in our inequality `y &gt; x^2 - 2`:

Left side is `5`

Right side is `1^2 - 2 = -1`

Is `5 &gt; -1`? Answer: yes!

Now let's check a point below our curve. Say `(6, 2)`. Substitute these values into our inequality and check it. It should fail.

Left side is `2`

Right side is `6^2 - 2 = 34`

Is `2 &gt; 34`? Answer: No.

So the region we should shade is the region above the curve.

Just another point... Your curve should be dotted because the question has `&gt;` sign, not `&gt;=`.

Now, you need to do the same process for the line `y = x` and decide if you should include the region above the line `y = x` or below.

X

okay i ve been able do something base on your instructions.am not sure if what i ve done is  okay. can u ve a look at the attached file and correct if possible.

<a href="/forum/uploads/double-integral.jpg"><img src="/forum/uploads/double-integral310.jpg" alt="double integral" width=310" height="361" /></a>

X

Hi Chris

Yay - you've got it now. Well done.

Only one picky comment. On your diagram you pointed to the parabola and wrote "y > x^2 - 2", and the 2 straight lines involved and wrote "x > 1" and "y < x". In fact, your labels should be `y = x^2 - 2` for the parabola and `x = 1` and `y = x` for the lines. It is the shaded region that should be labeled with the inequalities.

As for your double integral - yes, that is correct too. Do you know what your double integral actually means?

All the best.

X

thanks you are a good tutor.

actually double integral denotes vol of a surface

X

You're welcome, Chris. Glad to be of help. 

One final picky point - a double integral represents a volume below a surface bounded by an area (usually on the `x`-`y` plane). A surface can only have area, not volume.