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

The object's position is increasing on the interval t > 0, but it is not concave down at any point.

Step-by-step explanation:

To find the interval in which the object's position is increasing and concave down, we need to analyze the behavior of the first derivative and second derivative of the function. The first derivative of d(t) is given by d'(t) = (2t + 3)e^(-1/2), and the second derivative is given by d''(t) = 2e^(-1/2).

Since the first derivative is positive for any value of t, the object's position is increasing on the entire interval where the function is defined (t > 0).

The second derivative is always positive, indicating that the function is concave up for all values of t. Therefore, the object's position is not concave down at any point.

Answer 2

Step-by-step explanation:

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

so it is concave upward at t=0