A climber on Mt Everest is 6000 meters from the start of his trail and at elevation 8000 meters above sea level. At x meters from the start, the elevation is h(x) meters above sea level. If h′(x)=0.6 for x near 6000 meters, what is the approximate elevation another 5 meters along the trail? Chapter 2, Section 2.3, Supplementary Question 001a Since the derivative tells how fast a function is changing, use the derivative at a point to estimate values of the function at nearby points. For local linear approximation use Δh=h′(x)Δx. What are the values for Δx and h′(x)? Δx= meters and h′(x)= . Click if you would like to Show Work for this question: Open Show Work

We also know that the derivative of a function tells us the reason of change at some point (or how fast a function is changing near that point). For this problem we have that h'(x) = 0.6. (meaning that the elevation increases in 0.6 meters for every meter they hike further nearby the 6000 meters)

We need to use the derivative at a point to estimate values of the function at nearby points

We're going to use h'(x)= 0.6. Δx= 1

h(6001) = h(6000) + h'(6000)(1) = 8000 +0.6 = 8,000.6

Therefore h (6001) = 8,000.6

Now for h'(x)= 0.6. Δx= 1

h(6002) = h(6000) + h'(6000)(2) = 8000 + 0.6 (2) =8000 +1.2 = 8001.2

Therefore h(6002) = 8001.2