1. Part (a)
Remember that the 1st derivative is the slope. We can find the slope by using Algebra.
T’(7) =
8 - 6
T’(7) = -7 2
oC / cm

2. Part (b) 1 6 5 8
The average temperature would be represented by using the average value formula for the function… oC
½(1)(100+93)
½(4)(93+70)
½(1)(70+62)
½(2)(62+55)
Avg. temp = T(x) dx
??? 8 1
8-0 = 1 8
( )
96.5
326 66
117

3. According to the Fundamental Theorem,
Part (c)
According to the Fundamental Theorem,
??? 8
T’(x) dx =
T(8) - T(0)
??? 8
T’(x) dx =
= -45 oC
This represents the drop in temperature from one end of the wire to the other.

4. Part (d)
NO.
T”(x) > 0 implies that the 1st derivative is always increasing from x=0 to x=8.
From x=0 to x=1, T’(x) = -7
From x=1 to x=5, T’(x) = -23/4 (-5.75)
So far, so good…
From x=5 to x=6, T’(x) = -8
Since T’(x) is heading back down again, T”(x) cannot always be positive.