Integration (substitution) and inequality help (1 Viewer)

[blazer78]

Need help with question 2 and 3, i can't seem to finish it.

Heres what i've done so far for question 2:

i can only get one case to prove true, the other 2 seem dodgy...


and question 3:

am i going in the right direction?

 

[blazer78]

but theres only one point of intersection...


=====
for q3. i end up with: (e^x+3)(e^x-9)

am i right?

 

[pLuvia]

3. ∫ (2e^x)/(e^x+3) dx
Let u = e^x+3
du = e^x dx

.: 2 ∫ du/u
= 2ln(e^x+3) +C

 

[blazer78]

so the points of intersection are at x= -5/3 and x=2

lol i'm lost now...

 

[blazer78]

3. ∫ (2e^x)/(e^x+3) dx
Let u = e^x+3
du = e^x dx

.: 2 ∫ du/u
= 2ln(e^x+3) +C

lol, does that even involve substitution?

 

[blazer78]

ok, thanx i see how to do q3 now.

q2 is still giving me the creeps =(

 

[pLuvia]

You just have to manipulate the question. As 2 is a constant you can move it out of the integral sign which makes it easier to integrate

 

[blazer78]

through my method, i also found those 2, so the answer is [-7/2 , -3/4]

ie. -3/4 >= x

and -7/2 >= x

??

 

[blazer78]

You just have to manipulate the question. As 2 is a constant you can move it out of the integral sign which makes it easier to integrate

ok thanks!

 

[blazer78]

lol, thats alright, i should be thanking you for your efforts already =P

ok, so the answer is:

-7/2 =< x =< -3/4

?

 

[blazer78]

lol,

1.place sketch on paper in scanner.

2. turn on scanner

3. Run scanning software (mines called "EPSON Scan"

4. Click on "Scan"

5. Upload your image on imageshack "www.imageshack.us"

6. post a link to your image here

 

[blazer78]

Sorry, half of my post seems to have disappeared.

What I meant to say was, -7/2 < x < -3/4. I should have scanned my sketch and put it up earlier, but since I am a total technophobe, I don't actually know how to do that... (I'd be very happy if someone would enlighten me.)

well since the question has a greater than or EQUALS sign, shouldn't your answer be:

-7/2 =< x =< -3/4 ??

 

[blazer78]

if you look at my crude equational graph, you should find out how i get all 3 cases. and whether i made a mistake or not.

 

[blazer78]

lol it works! yay!

lol, i'm still wondering if the correct answer is

-7/2 =< x =< -3/4

or [ - infinity , -7/2] - interval notation ie.. negative infinity =< x =< -7/2

 

[blazer78]

ok solved.

If you look at my working out, case 3 is a contradiction, since if x = -7/2, for that case
-7/2 cannot be greater than 2

therefore, no solution for case 3.

case 2 and case 1 work out

therefore solution is

[-7/2 , -3/4] - interval notation

 

[blazer78]

thanks very much for your help Iruka!

 

[icycloud]

Just square both sides mate, much easier.

(x-2)^2 - (3x+5)^2 >= 0
x^2 - 4x + 4 - 9x^2 - 30x - 25 >= 0
-8x^2 - 34x - 21 >= 0
8x^2 + 34x + 21 <= 0
(x+3/4)(x+7/2) <= 0

Thus, -7/2 <= x <= -3/4
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