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

Explanation:

Step1: Find the derivative of the displacement function

The displacement function is $s(t)=7 - 5t+\frac{5t^{3}}{192}$. Using the power - rule for differentiation $\frac{d}{dt}(t^n)=nt^{n - 1}$, we have $s^\prime(t)=\frac{d}{dt}(7)-\frac{d}{dt}(5t)+\frac{d}{dt}(\frac{5t^{3}}{192})$. Since $\frac{d}{dt}(7) = 0$, $\frac{d}{dt}(5t)=5$, and $\frac{d}{dt}(\frac{5t^{3}}{192})=\frac{5\times3t^{2}}{192}=\frac{5t^{2}}{64}$, then $s^\prime(t)=\frac{5t^{2}}{64}-5$.

Step2: Set the derivative equal to zero to find critical points

Set $s^\prime(t) = 0$. So, $\frac{5t^{2}}{64}-5 = 0$. Add 5 to both sides: $\frac{5t^{2}}{64}=5$. Multiply both sides by $\frac{64}{5}$ to get $t^{2}=64$. Take the square - root of both sides: $t=\pm8$. Since $t$ represents time and $0\leq t\leq10$, we consider $t = 8$.

Step3: Determine intervals of increase and decrease

We test the intervals $[0,8)$ and $(8,10]$. For $t\in[0,8)$, let's choose $t = 0$. Then $s^\prime(0)=\frac{5\times0^{2}}{64}-5=- 5<0$, so $s(t)$ is decreasing on $[0,8)$. For $t\in(8,10]$, let's choose $t = 9$. Then $s^\prime(9)=\frac{5\times9^{2}}{64}-5=\frac{405}{64}-5=\frac{405 - 320}{64}=\frac{85}{64}>0$, so $s(t)$ is increasing on $(8,10]$.

Answer:

Displacement is decreasing over: $[0,8)$ Displacement is increasing over: $(8,10]$