Answer
$\frac{dT}{dh}|_{h=30}$ = $2.94$ $°C/km$
$\frac{dT}{dh}|_{h=70}$ = $-3.33$ $°C/km$
$\frac{dT}{dh}=0$ is h = 50 and h = 90
Work Step by Step
h = 30 km, line passing through the points (23, -50) and (40, 0)
$\frac{dT}{dh}|_{h=30}$ = $\frac{0-(-50)}{40-23}$ = $2.94$ $°C/km$
h = 70 km, line passing through the points (58, 0) and (88, -100)
$\frac{dT}{dh}|_{h=70}$ = $\frac{0-(-100)}{58-88}$ = $-3.33$ $°C/km$
$\frac{dT}{dh}=0$ where the tangent line on the graph is horizontal, it is h = 50 and h = 90
