- #1
RJLiberator
Gold Member
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1) True or Fale?
|a-b| [itex]\geq[/itex] ||a|-|b||
My solution: I broke this up into 4 different cases. 1. a> 0 b>0 2. a<0 b<0 and so on...
For each case, I ended up with a more simplified version of the inequality. For instance, in case 1 where I used a<0 and b<0, I ended up with the simplified statement of:
[itex]|-a+b|\geq|-a+b|[/itex] Which is clearly equivalent.
Is this a correct way to prove that this statement is true?
2. With that being out of the way, how can I prove this statement to be true:
[itex]\sqrt{(a+b)^2} = |a+b|[/itex]
Would I do the same thing and break it up into 4 cases?
3. This has been my most difficult problem so far.
For any x > 0, consider the function
f(x)=[itex]\int^{\sqrt{x}}_{0}e^{t^{2}}dt[/itex]
Computer f'(x)
My work so far: I was told to use the chain rule where the inner function is [itex]\sqrt{x}[/itex] and the outer is 0.
I'm not qutie sure what to do.. I imagine I take the antiderivative of this function using the chain rule?
|a-b| [itex]\geq[/itex] ||a|-|b||
My solution: I broke this up into 4 different cases. 1. a> 0 b>0 2. a<0 b<0 and so on...
For each case, I ended up with a more simplified version of the inequality. For instance, in case 1 where I used a<0 and b<0, I ended up with the simplified statement of:
[itex]|-a+b|\geq|-a+b|[/itex] Which is clearly equivalent.
Is this a correct way to prove that this statement is true?
2. With that being out of the way, how can I prove this statement to be true:
[itex]\sqrt{(a+b)^2} = |a+b|[/itex]
Would I do the same thing and break it up into 4 cases?
3. This has been my most difficult problem so far.
For any x > 0, consider the function
f(x)=[itex]\int^{\sqrt{x}}_{0}e^{t^{2}}dt[/itex]
Computer f'(x)
My work so far: I was told to use the chain rule where the inner function is [itex]\sqrt{x}[/itex] and the outer is 0.
I'm not qutie sure what to do.. I imagine I take the antiderivative of this function using the chain rule?
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