> f'(x) is negative for all x, and f''(x) always has the same sign as x.
Say, let f'(x) = -e^{-x^{2}}, and f''(x) = 2xe^{-x^{2}};
f(x) = \Int{f'(x)dx} = -0.5sqrt(pi)erf(x) + C, where erf is the error function(, OK, I cheated with Wolfram Alpha, and never worked out the integral part myself).
But the point is to draw a function with these properties. You just have to have a smooth curve that approaches 0 asymptotically and is concave up. No worries about graphing any particular function!
Either you got the description wrong or I'm especially rusty - in that case, f''(x) = e^(-x), which is positive even when x is negative, so it doesn't always have the same sign.
Say, let f'(x) = -e^{-x^{2}}, and f''(x) = 2xe^{-x^{2}}; f(x) = \Int{f'(x)dx} = -0.5sqrt(pi)erf(x) + C, where erf is the error function(, OK, I cheated with Wolfram Alpha, and never worked out the integral part myself).