4 ms·
> 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} = -0.5sq
by ttflee 13y ago
> 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).
- wging 13y agof(x) = e^(-x) works too :) 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!
- jackmaney 13y agoEven easier: f(x) = - arctan(x). We have f'(x) = -1/(1+x^2) < 0 for all x, and f''(x) = 2*x/((1+x^2)^2) which has the same sign as x.