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This is one of our many quizzes on Graphing logistic functions. Click on this to access more such competitive quizzes on Algebra.

Graphing logistic functions Quiz 2

mathematics

 A hospital is using the logistic function $f(t)=\dfrac{16}{1+2{{e}^{-t}}}$. Calculate the value of $t$such that $f(t)=1$ and identify the y-intercept of the function. 

 Substitute the value of $f(t)=1$ and solve for $t$. For calculating the $y-$ intercept, substitute the value of $x=0$ and solve for $f(0)$. 

  $1=\dfrac{16}{1+2{{e}^{-t}}} \\ 1+2{{e}^{-t}}=16 \\ 2{{e}^{-t}}=15 \\ {{e}^{-t}}=\dfrac{15}{2} \\ \log ({{e}^{-t}})=\log \left( \dfrac{15}{2} \right) \\ t=-\log \left( \dfrac{15}{2} \right) \\ f(0)=\dfrac{16}{1+2{{e}^{-0}}} \\ f(0)=\dfrac{16}{3}$

 Using two points, $L=400,(0,75),(2,45)$, calculate the logistic model given that the value of $L$is $400$. 

 Substitute one point at a time such that you can yield the value of one of the variables. You should then substitute the value with the second point.

  $f(x)=\dfrac{C}{1+a{{b}^{x}}} \\ 75=\dfrac{400}{1+a{{b}^{0}}} \\ 75=\dfrac{400}{1+a} \\ a=4.\overline{3} \\ 45=\dfrac{400}{1+4.\overline{3}{{b}^{2}}} \\ 1+4.\overline{3}{{b}^{2}}=\dfrac{400}{45} \\ {{b}^{2}}=\dfrac{\dfrac{355}{45}}{4.3} \\ b=\sqrt{\dfrac{355}{193.5}} \\ f(x)=\dfrac{400}{1+\dfrac{13}{3}{{\left( \sqrt{\dfrac{355}{193.5}} \right)}^{x}}}$

 At a given school, the population of the number of students is $S(t)$. A statistician calculated that the student population $S(t)$ invariably satisfies the logistic differential equation $\dfrac{dS}{dt}=4S\times \left( 4-\dfrac{S}{6000} \right)$ such that the variable $t$ is used to measure the population in years. The initial population is $400$ students. Calculate the carrying capacity of the school. 

 Multiply both sides by a constant such that the differential equation can come to the form $\dfrac{dS}{dt}=kS(a-S)$ where $a$ is the carrying capacity

  $\dfrac{dS}{dt}=kS(a-S) \\ a= \text{ Carrying Capacity } \\ \text{Multiply both LHS and RHS with }\dfrac{6000}{6000} \\  \dfrac{dS}{dt}\times \dfrac{6000}{6000}=4S\times \left( 4-\dfrac{S}{6000} \right)\times \dfrac{6000}{6000} \\ \dfrac{dS}{dt}=4S\times \dfrac{1}{6000}\times \left( 4-\dfrac{S}{6000} \right)\times 6000 \\ \dfrac{dS}{dt}=\dfrac{4S}{6000}\times \left( 4\times 6000-S \right) \\ \dfrac{dS}{dt}=\dfrac{S}{1500}\times \left( 24000-S \right) \\ \text{Therefore, 24000 is the carrying capacity.}$

 A housing society has the number of residents, $R$ such that the total number of residents are decreasing. A logistic function can be modelled decaying using the function $f(x)=400{{(0.4)}^{x}}$ where $x$ represents the number of years since the highest recorded resident count in the society. Calculate after how many years will there be just 100 residents given that $\ln (2)=0.693$ and $\ln (0.4)=-0.916$? 

 Substitute the value of the number of residents and solve for $x$

  $100=400{{(0.4)}^{x}} \\ \dfrac{100}{400}={{0.4}^{x}} \\ \dfrac{1}{4}={{0.4}^{x}} \\ \ln \left( \dfrac{1}{4} \right)=\ln ({{0.4}^{x}}) \\ x\ln (0.4)=\ln (1)-\ln (4) \\ -0.916x=0-2\ln (2) \\ -0.916x=0-2(0.693) \\ x=1.513 $

 Calculate the x-coordinate for the inflection point of the function, $f(x)=\dfrac{1}{1+{{e}^{-x}}}$.

 Calculate the double derivative of the function in order to find the inflection point

  The logistic function doesn’t intersect the x-axis, instead, it touches it, hence, there is no x-intercept. For calculating the inflection, calculate the double derivative of the logistic function, $ \\ \dfrac{d}{dx}\left( \dfrac{1}{{{e}^{-x}}+1} \right)=\dfrac{-\dfrac{d}{dx}\left( {{e}^{-x}}+1 \right)}{{{\left( {{e}^{-x}}+1 \right)}^{2}}} \\ -\dfrac{\dfrac{d}{dx}\left( {{e}^{-x}} \right)+\dfrac{d}{dx}\left( 1 \right)}{{{\left( {{e}^{-x}}+1 \right)}^{2}}}=\dfrac{{{e}^{-x}}\cdot \dfrac{d}{dx}(-x)}{{{\left( {{e}^{-x}}+1 \right)}^{2}}} \\ \dfrac{{{e}^{-x}}}{{{\left( {{e}^{-x}}+1 \right)}^{2}}}=\dfrac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}} \\ f''(x)=\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}} \right) \\ \dfrac{\dfrac{d}{dx}\left( {{e}^{x}} \right)\cdot {{\left( {{e}^{x}}+1 \right)}^{2}}-{{e}^{x}}\cdot \dfrac{d}{dx}{{\left( {{e}^{x}}+1 \right)}^{2}}}{{{\left( {{\left( {{e}^{x}}+1 \right)}^{2}} \right)}^{2}}} \\ \Rightarrow \dfrac{{{e}^{x}}{{\left( {{e}^{x}}+1 \right)}^{2}}-2\left( {{e}^{x}}+1 \right){{e}^{2x}}}{{{\left( {{e}^{x}}+1 \right)}^{4}}} \\ \dfrac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}-\dfrac{2{{e}^{2x}}}{{{\left( {{e}^{x}}+1 \right)}^{3}}} \\ \Rightarrow \dfrac{{{e}^{-2x}}(-{{e}^{x}}+1)}{{{\left( {{e}^{-x}}+1 \right)}^{3}}} \\ $ $f''(x)$ is undefined at $x=0$, Therefore $x=0$