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Calculus help please

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True or false: there is a function, defined for all real numbers, such that f(x)>0, f'(x)>0, and f''(x)<0 for all real x.

True
False

I'm thinking it is true.
Posted: October 2013 Permalink

Replies

you can always test it can you not? idk if that helps tho

f(x)=x^3 + 2x^2 + x
f'(x) = 3x^2+4x+1
f''(x)=6x+4

in this case it looks true. forgive me if im wrong this was like the last lesson in gr 12 calc before i dropped it ='(
Oct 27 2013
goatman0079 Level 152 Arcania Night Walker 4
it could be true if f(x) = sin(x) because the deriv of sin(x) = cos(x) and the second deriv of sin(x) = -sin(x)
Oct 27 2013
cb000 Level 222 Bera Hero
False. A concave down function will either arch down towards the x axis or have arched from the xaxis to some asymptotic ceiling. In either case it must cross the x-axis somewhere, assuming it must be defined for all real X from -inf to inf.

@Above: none of those functions is strictly positive or strictly negative for all x.

Edit: Actually, would a periodic, piecewise function work? Like having sin(x) for pi/6<=x<pi/3 repeated throughout. It would be discontinuous, but yea.
Oct 27 2013
goatman0079 Level 152 Arcania Night Walker 4
Actually it seems like it would be false because if the derivative is >0, then F(x) would have to be negative at some point
Oct 27 2013

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