Rate of Change Problem: Finding Derivative of Temperature Function T(t)

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Incog
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Homework Statement



Suppose that t hours after a piece of food is put in the fridge its temperature (in Celsius) is

T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

where 0 [tex]\leq[/tex] t [tex]\leq[/tex] 5.

Find the rate of change of temperature after one hour.


The Attempt at a Solution



Since it's asking for rate of change, I'm guessing I have to find the derivative of the equation with respect to t.


T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

T`(t) = 0 - 3 + [tex]\frac{0(t - 1) - 1(4)}{(t-1)^{2}}[/tex] (Quotient Rule)

T`(t) = -3 + [tex]\frac{0 - 4}{(t-1)^{2}}[/tex]

T`(t) = -3 + [tex]\frac{-4}{(t-1)^{2}}[/tex]

T`(t) = -3 - [tex]\frac{4}{(t-1)^{2}}[/tex]


Would I just plug in 1 after this?
 
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Incog said:

Homework Statement



Suppose that t hours after a piece of food is put in the fridge its temperature (in Celsius) is

T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

where 0 [tex]\leq[/tex] t [tex]\leq[/tex] 5.

Find the rate of change of temperature after one hour.


The Attempt at a Solution



Since it's asking for rate of change, I'm guessing I have to find the derivative of the equation with respect to t.
Don't guess! The derivative of a function is its rate of change!


T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

T`(t) = 0 - 3 + [tex]\frac{0(t - 1) - 1(4)}{(t-1)^{2}}[/tex] (Quotient Rule)

T`(t) = -3 + [tex]\frac{0 - 4}{(t-1)^{2}}[/tex]

T`(t) = -3 + [tex]\frac{-4}{(t-1)^{2}}[/tex]

T`(t) = -3 - [tex]\frac{4}{(t-1)^{2}}[/tex]


Would I just plug in 1 after this?
That's what you would like to do- but this function has serious problem at t= 1. Do you remember that, in order to have a derivative at a point, the function must be continuous there? Are you sure you have copied the problem correctly? That's a very strange temperature function! Isn't it peculiar that the temperature of the food goes up when it is put in the refridgerator?
 
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Yes, I checked and checked again and that is the equation.

What if I were to plug in a value slightly greater than 1? Would that give me the rate of change after one hour?
 
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