#### To determine

**To find**: Inflection point of S.

#### Answer

t=28.57 and t=85.71

#### Explanation

**Given**: S(t)=Atpe−kt.

S′(t)=A((tp)(−ke−kt)+(e−kt)(ptp−1)) =Atpe−kt(pt−k)=S(t)(pt−k).

S″(t)=S′(t)(pt−k)+S(t)(−pt2)=S(t)(pt−k)(pt−k)+S(t)(−pt2)=Atpe−kt((pt−k)2−pt2)=Atpe−kt(p2t2−2pkt+k2−pt2)=Atpe−kt(p2−pt2−2pkt+k2). Now, S″(t)=0 if (p2−pt2−2pkt+k2)=0. This equation is quadratic in 1t, therefore

1t=2pk±(−2pk)2−4k2(p2−p) 2(p2−p)=pk±kp p2−p. That means, t=p2−ppk±kp, i.e., t=p2−ppk+kp, and t=p2−ppk−kp. Substitute k=0.07, p=4, we get t=28.57 and t=85.71.

**Conclusion: **t=28.57 and t=85.71