The extreme value theorem applies to closed and bounded intervals only. The interval in your question is not closed, so the extreme value theorem doesn't apply here. You can see what's going on by sketching the graphs of those functions on the given interval.Another question, but mainly to do with the fact that I haven't been fully taught how to apply the extreme value theorem yet
Fair enough. I knew I should've believed that it has to be a closed interval.The extreme value theorem applies to closed and bounded intervals only. The interval in your question is not closed, so the extreme value theorem doesn't apply here. You can see what's going on by sketching the graphs of those functions on the given interval.
Sorry guys, have to bump this one because this is now screwing my head around a bit too muchDisappointing. What would be the inverse to:
f: R->R, f(x) = x^3
then, if g(x) = x^(1/3) is undefined for x<0?
The inverse is simply the function g: R -> R given by g(x) = cube root of x, where by cube root we mean the real cube root.Sorry guys, have to bump this one because this is now screwing my head around a bit too much
x^{1/3} is perfectly well defined for x<0, even if you have never heard of complex numbers.Sorry guys, have to bump this one because this is now screwing my head around a bit too much
That's what I thought until Wolfram and GeoGebra messed up my brain.x^{1/3} is perfectly well defined for x<0, even if you have never heard of complex numbers.
x^3 is a bijection from R to itself.
For Q3, a direction vector for the line is the vector that the lambda is multiplying in your parametric form (that's the direction the line points in).The answers for these are not uploaded, so I am requesting a marking service. (Sorry in advance as this is probably wasting people's time now.) Also some help on Q3 because I didn't quite understand it.
But to save space (and time) I'll drop out a lot of working out and just give the final results
Q1
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Q2
Note: The answer isn't concerned with a diagram being to scale so my triangle was arbitrarily drawn, just with points labelled
Answer: The entire region bounded by triangle ABC, where A is (1,2,-2), B is (3,-2,-2) and C is (1,-8,0)
Method: The boundary cases are when lambda1=lambda2=0, lambda1=2, lambda2=0 and lambda1=lambda2=2
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Q3
Answer: I don't know what a direction vector is anymore so I just converted...
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Q4
Row-echelon form:
Answer:
a) m=8 otherwise contradiction in row 3
b) never, all variables are leading variables provided m can't equal 8
c) m \neq 8
...ohFor Q3, a direction vector for the line is the vector that the lambda is multiplying in your parametric form (that's the direction the line points in).
Yeah....oh
So it's just (0,-1,5)?
Let eps > 0. Take delta = min(2, eps/4). Let 0 < |x - 1| < delta. Then |x-1| < eps/4. Also, |x-1|<2, so -1 < x < 3, whence 0 < x+1 < 4. So |x+1| < 4.I still don't fully know what I'm doing. Another demo please, but on a trivial statement this time
All I can say is
I have no clue on the method
Sorry, forgot what that meant againTake delta = min(2, eps/4).
Delta is the smaller of the two. The importance of this is that if we make something less than delta, it'll be simultaneously less than 2 and eps/4.Sorry, forgot what that meant again