Find the exact location of all the relative and absolute extrema of the function g(t) = 12e^−t^2

Question

domain of (negative infinity, positive infinity)

Answer 1

g(t)=12e^-t²,

g'(t)=-24te^-t² is zero at an extremum, that is, when t=0, the point (0,12).

g''(t)=-24e^-t²+48t²e^-t².  g''(0)=-24. This is negative concavity so there is a maximum at (0,12).

As t→±∞, g(t)→0. The t-axis is the asymptote. (0,12) is also an absolute maximum. There is no absolute minimum because g(t) cannot be equal to zero.