Page 81 - Calculus Demystified

CHAPTER 2
Foundations of Calculus
68 by (c, f (c)) and nearby points (c + h, f (c + h)). See Fig. 2.8. Let us calculate this
limit:
f(c + h) − f(c) f(c + h) − f(c)
lim
h→0 (c + h) − c = lim h .
h→0

Fig. 2.8
We know that this last limit (the same as (∗)) is the derivative of f at c. We have
learned the following:
Let f be a differentiable function on an interval (a, b). Let c ∈ (a, b). Then
the slope of the tangent line to the graph of f at c is f (c ).

EXAMPLE 2.11 TEAMFLY
Calculate the instantaneous velocity at time t = 5 of an automobile whose
position at time t secondsisgiven by g(t) = t + 4t + 10 feet.
3
2
SOLUTION
We know that the required instantaneous velocity is g (5). We calculate

g (5) = lim g(5 + h) − g(5)

h→0 h
3
2
2
3
= lim [(5 + h) + 4(5 + h) + 10]−[5 + 4 · 5 + 10]
h→0 h
((125 + 75h + 15h + h ) + 4 · (25 + 10h + h ) + 10)
2
2
3
= lim h
h→0

− (125 + 100 + 10)
h
115h + 19h + h 3
2
= lim h
h→0
= lim 115 + 19h + h 2
h→0
= 115.

Team-Fly
®