Assume the function f is such that f'' (x) > 0 for all values of x. Which one of following statement is true about f?
see picture
A can't be right think of \(e^x\)
I know it's not A or B
i guess it can't be C either,again think of \(e^x\) , no zeros, no critical point
and similarly D can''t be right, for the same reason that leave E
of course if the critical number exists, and if \(f''>0\) this means it must be a local min i guess it is also true that there can be at most one, since if \(f''>0\) that means \(f'\) is always increasing, so it cannot be zero more then once
how do u know that f behaves like e^x ?
the only info u have is f is concave up.. but it still can mean it crosses the xintercept right @satellite73
E is the correct ans..see,, it is given that rate of change of slope is always positive.. which means slope must be negative somewhere and is increasing as x increasing, hence it'll go from -ve to 0 to +ve,,which means minima..
I do understand there is a minima, and slope is always increasing oh wait, i get it since there is a minima, that means f'(x) = 0 , which means 1 critical point , is that right?
no one says it crosses the \(x\) axis
E says it has AT MOST one critical point, and IF it has one, it is a min
yea, i'm keep mixing up root and ciritical pt :/
again \(e^x\) is a good example it has not critical point, and never crosses the \(x\) axis it was an "if" statement
btw the critical point if it exists must be root of the derivative, it cannot be where the derivative is undefined writing \(f''<0\) assumes that \(f''\) exists for all \(x\) and therefore \(f'\) exists for all \(x\) as well
@satellite73 hey, just a quck quesion. If f'' exist, f' cant be undefined rght
yes that is right if \(f''\) exists then \(f'\) must not only be defined, it must be continuous as well
ok thanks!
now i have a question
what drink are you making?
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