Monday 13 March 2017

The yearly profit of a company in thousands of dollars can be modeled by P(t) = t4 − 5t2 + 4, where t is the number of years since...

We are given `P(t)=t^4-5t^2+4 ` as a model for the profits earned, where t is the time in years since 2005. We are asked to find the years with zero profit.


This is equivalent to solving the equation `t^4-5t^2+4=0 ` .


Note that this is a quadratic form -- it is a quadratic in `t^2 ` .


`t^4-5t^2+4=(t^2-4)(t^2-1) `


Each of the binomials is a difference of two squares and can be factored as:


`=(t+2)(t-2)(t+1)(t-1)...

We are given `P(t)=t^4-5t^2+4 ` as a model for the profits earned, where t is the time in years since 2005. We are asked to find the years with zero profit.


This is equivalent to solving the equation `t^4-5t^2+4=0 ` .


Note that this is a quadratic form -- it is a quadratic in `t^2 ` .


`t^4-5t^2+4=(t^2-4)(t^2-1) `


Each of the binomials is a difference of two squares and can be factored as:


`=(t+2)(t-2)(t+1)(t-1) `


So (t+2)(t-2)(t+1)(t-1)=0


`t=pm1,pm2 `


Since the domain is positive (the model is true for years after 2005), t=1 is 2006 and t=2 is 2007.


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The profit was zero in 2006 and 2007


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There were 4 solutions, but 2 of them (-1 and -2) were not in the domain and so are disregarded. (If the company came into existence in 2005, it makes no sense to describe its profits before 2005.)


The graph for t>0:


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