the displacement of a bison measured from a geyser (in feet) is given by s(t)=7 - 5t+\\frac{5t^{3}}{192}…

the displacement of a bison measured from a geyser (in feet) is given by s(t)=7 - 5t+\\frac{5t^{3}}{192}, where t represents the number of seconds after the bison starts moving. the bison only moves for 10 seconds. over what intervals of time is the bisons displacement from the geyser decreasing and increasing? enter dne if no such interval exists. displacement is decreasing over: displacement is increasing over:

the displacement of a bison measured from a geyser (in feet) is given by s(t)=7 - 5t+\\frac{5t^{3}}{192}, where t represents the number of seconds after the bison starts moving. the bison only moves for 10 seconds. over what intervals of time is the bisons displacement from the geyser decreasing and increasing? enter dne if no such interval exists. displacement is decreasing over: displacement is increasing over:

Answer

Explanation:

Step1: Find the derivative of (s(t))

We have (s(t)=7 - 5t+\frac{5t^{3}}{192}). Using the power - rule ((x^n)^\prime=nx^{n - 1}), the derivative (s^\prime(t)=-5+\frac{15t^{2}}{192}=-5+\frac{5t^{2}}{64}).

Step2: Set (s^\prime(t)=0) to find critical points

[ \begin{align*} -5+\frac{5t^{2}}{64}&=0\ \frac{5t^{2}}{64}&=5\ t^{2}&=64\ t&=\pm8 \end{align*} ] Since (t) represents time and the bison moves for (0\leq t\leq10), we consider (t = 8) (we ignore (t=-8)).

Step3: Test intervals using the first - derivative test

We consider the intervals ([0,8)) and ((8,10]). For (t\in[0,8)), let's choose (t = 1). Then (s^\prime(1)=-5+\frac{5}{64}<0). For (t\in(8,10]), let's choose (t = 9). Then (s^\prime(9)=-5+\frac{5\times81}{64}=\frac{- 320 + 405}{64}=\frac{85}{64}>0).

Answer:

((8,10])