Differentiation and continuity problems–221 fall
2015
Oct 5 2015
Differentiate the following using the definition of derivative:
1
f (x) =
1
, x 6= 0
x
1
−
f (x + h) − f (x)
f (x) = lim
= lim x+h
h→0
h→0
h
h
−1
−1
= 2
= lim
h→0 x(x + h)
x
0
1
x
= lim
hto0
x−(x+h)
x(x+h)
h
2
f (x) = 4 −
√
x+3
√
√
f (x + h) − f (x)
4 − x + h + 3 − (4 − x + 3)
0
f (x) = lim
= lim
=
h→0
h→0
h
h
√
√
x+3− x+h+3
(x + 3) − (x + h + 3)
√
lim
= lim √
h→0
h→0 h( x + 3 +
h
x + h + 3)
−1
−1
√
= lim √
= √
h→0
x+3+ x+h+3
2 x+h+3
3
f (x) = x2/3
f (x) − f (a)
x2/3 − a2/3
x2 − a2
= lim
= lim
4/3
x→a
x→a
x→a (x − a)(x
x−a
x−a
+ x2/3 a2/3 + a4/3 )
f 0 (a) = lim
= lim
x→a
x+a
2
= a−1/3
3
x4/3 + x2/3 a2/3 + a4/3
1
4
a and b are real numbers so that the following function is continuous on all of
R. What are a and b?

:x≤0
 sin(ax) + b
bx2 + a
:0<x<1
f (x) =

2
:x≥1
We know for a function to be continuous at a is exactly for the following to
be true:
lim f (x) = f (a)
x→a
This is equivalent to
lim f (x) = lim+ f (x) = f (a)
x→a−
x→a
Applying this to the above (noticing we only have to worry about 0 and 1),
we see to make our function continuous at 0, we know a and b must satisfy:
lim f (x) = lim+ f (x) = f (0)
x to0−
x→0
b=a=b
by computing the side limits. Similarly, at 1, we have
lim f (x) = lim f (x) = f (1)
x to1−
x→1+
b+a=2=2
so we have 2 equations and 2 unknowns. Solve ’em to get a = b = 1.
5
a and b are real numbers so that the following function is continuous and differentiable on all of R. What are a and b?
sin(ax) : x ≤ 0
f (x) =
x+b
:x>0
Here, first we want to make this function continuous at 0. Proceeding as
before:
lim f (x) = lim+ f (x) = f (0)
x to0−
x→0
0=b=0
But we don’t know what a is yet. This is where we’re going to make it differentiable.
2
Remember being differentiable means the limit if your slopes exists. Again,
we only need to worry about 0. That is, being differentiable at 0 is the same
thing as this limit existing:
lim
h→0
f (0 + h) − f (0)
h
Which is the same thing as the side limits existing and being equal. So we need
lim−
h→0
f (0 + h) − f (0)
f (0 + h) − f (0)
= lim+
h
h
h→0
But what are these things? If h < 0 we’re only looking at sin(ah). If h > 0
we’re only looking at h + b. That is:
lim
h→0−
sin(a(0 + h)) − sin(0)
h
= lim+
h
h→0 h
Here the limit on the right is really easy to compute but in general, we know that
the function that f is when it’s bigger than 0 is differentiable at 0, so the limit
of the slopes from the right is the same thing as the value of the derivative there,
by the definition of being differentiable! Similarly, the limit of those slopes on
d
sin(ax)(0). That is, the derivative of sin(ax)
the left is the same thing as dx
evaluated at the point 0. So we see the above equality is
a cos(a × 0) = 1 ⇒ a = 1
So we have
a=1
b=0
Differentiate the following by any means necessary:
6
f (x) = 2x sin(x)
0
f (x) = 2x cos(x) + 2 sin(x)
7
f (x) = 33x3
f 0 (x) = 99x2
3