function bounded on an interval?

we have a function

f(x) = x^2 - 3 for x<2, 6/x for x>=2 (x greater or equal to 2)

i've shown that the limits of this function from the left and right of 2 respectively are 1 and 3. (given in qn)

also shown that ***if any function is continuous on [a,b] then it is bounded on [a,b]***

now i need to find out whether i can

1) use the above theorem (in bold) to the function f(x) on the interval [0,5]

2) determine whether f(x) is bounded on [0,5]

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my thoughts - clearly the function isn't continuous at 2, but it is piecewise continuous on the interval, so.... it's okay to use the theorem? or does it have to be continuous everywhere?

apologies for the lack of latex

thank youuu

Re: function bounded on an interval?

Quote:

Originally Posted by

**cassius**

also shown that ***if any function is continuous on [a,b] then it is bounded on [a,b]***

apologies for the lack of latex

thank youuu

It is Weierstrass first theorem.(from Hebrew)

Extreme value theorem - Wikipedia, the free encyclopedia

Re: function bounded on an interval?

Quote:

Originally Posted by

**cassius** we have a function

f(x) = x^2 - 3 for x<2, 6/x for x>=2 (x greater or equal to 2

also shown that ***if any function is continuous on [a,b] then it is bounded on [a,b]***

now i need to find out whether i can

1) use the above theorem (in bold) to the function f(x) on the interval [0,5]

2) determine whether f(x) is bounded on [0,5]

1) Because is not continuous on you cannot use that theorem.

2) But you can prove it is bounded there any simple inequalities.

Re: function bounded on an interval?

if we draw the graph of f(x) we can see it is bounded below by -3 at x=0; is this correct?

thank you

Re: function bounded on an interval?

Quote:

Originally Posted by

**cassius** if we draw the graph of f(x) we can see it is bounded below by -3 at x=0; is this correct?

if

Re: function bounded on an interval?