Chapter 13 - Section 13.3 - Tangent Lines and Derivatives - 13.3 Exercises - Page 923: 41

$-0.78^{\circ}$F/ min

Use the figure given by the Exercise, the rate of change of the temperature after 1 hour equals to the slope of the tangent line at the point $(60, 125)$. We can also identify a point $(15,160)$ on the tangent line. Thus the slope is $m=\frac{125-160}{60-15}\approx-0.78^{\circ}$F/ min