5 Steps to a 5 AP Calculus BC 2019

Differentiation 77

x

f(x)

x = a

a

f is discontinuous

at x = a

0

y

Figure 6.1-2

f has a corner

at x = a

y

x

f(x)

(a, (f(a))

(^0) a
Figure 6.1-3
f has a cusp at x = a
y
x
a^
(a, (f(a))
f(x)
0
Figure 6.1-4
f(x)
x=a
f has a vertical
tangent at x = a
y
x
(a, (f(a))
0 a
Figure 6.1-5
Example 1
If f(x)=x^2 − 2 x −3, find (a) f′(x) using the definition of derivative, (b) f′(0),
(c)f′(1), and (d) f′(3).
(a) Using the definition of derivative,f′(x)=limh→ 0
f(x+h)−f(x)
h
=limh→ 0
[(x+h)^2 −2(x+h)−3]−[x^2 − 2 x−3]
h
=limh→ 0
[x^2 + 2 xh+h^2 − 2 x− 2 h−3]−[x^2 − 2 x−3]
h
=limh→ 0
2 xh+h^2 − 2 h
h
=limh→ 0
h(2x+h−2)
h
=limh→ 0 (2x+h−2)= 2 x− 2.
(b) f′(0)=2(0)− 2 =−2, (c) f′(1)=2(1)− 2 =0 and (d)f′(3)=2(3)− 2 =4.