Question

If an object's position over time is given by the function d(t)=(t^2 +3)e^-1/2 for greater than 0 , on what interval is the object's position increasing and concave down ?

Answer 1

Answer:

Step-by-step explanation:

$d(t)=(t^2+3)e^\frac{-1}{2} \\d'(t)=2t e^{-\frac{1}{2} }\\it~ is~ increasing ~if~d'(t)>0\\e^{-\frac{1}{2} } =\frac{1}{e^\frac{1}{2} } >0\\so~t>0\\and~decreasing~if~t<0$

so it is concave upward at t=0