MATH 100 V1A
September 17th – Practice Problems
Solutions
Note: For all of the questions below, derivatives should be calculated using the limit definition
rather than any differentiation rules you may already know.
1. Find the equation of the tangent line to y = 3x2 + 2x at x = 1. Then sketch both
curves.
Solution: Let f (x) = 3x2 + 2x. Then
f (1 + h) − f (1)
3(1 + h)2 + 2(1 + h) − 5
= lim
h→0
h→0
h
h
3(1 + 2h + h2 ) + 2 + 2h − 5
= lim
h→0
h
8h + 3h2
= 8,
= lim
h→0
h
f 0 (1) = lim
so the slope of the line tangent to f at x = 1 is 8. Therefore, the equation of the line
tangent to f at x = 1 is given by
y − 5 = 8(x − 1) or y = 8x − 3.
Both curves are graphed below.
8 y
6
4
2
x
−2
2
4
6
2. Let
(
f (x) =
1
,
x
0,
if x 6= 0
.
if x = 0
Where does f 0 (x) not exist?
Solution: First let us consider the case when x 6= 0. Since we’re concerned with the
limit as h goes to 0, we may assume that h is close enough to 0 so that x + h 6= 0.
Then we have that
f (x + h) − f (x)
= lim
h→0
h→0
h
lim
= lim
1
x+h
− 1x
h
x−(x+h)
x(x+h)
h
−1
1
= lim
= − 2.
h→0 x(x + h)
x
h→0
So, for x 6= 0, we see that f 0 (x) = − x12 .
Now, let us consider the case when x = 0. We have that
f (0 + h) − f (0)
1
= lim 2 ,
h→0
h→0 h
h
lim
which does not exist (+∞), so f is not differentiable at x = 0.
3. Let
(
−x, if x < 0
f (x) =
,
x,
if x ≥ 0
and let g(x) = xf (x). Calculate g 0 (x) and g 00 (x). Are there any points where g 0 (x) or
g 00 (x) do not exist?
Solution: Let us first consider the case when x > 0. Then, since we considering the
limit as h goes to 0, we may assume that h is close enough to 0 so that x + h > 0.
Then we have that
g(x + h) − g(x)
(x + h)2 − x2
= lim
h→0
h→0
h
h
2
x + 2xh + h2 − x2
= lim
= 2x.
h→0
h
lim
So, for x > 0 we have that g 0 (x) = 2x. A similar calculation shows that g 0 (x) = −2x
when x < 0. Now let us consider the case when x = 0. If we calculate the right-hand
limit (where we only consider h > 0, denoted limh→0+ ) we get
lim+
h→0
g(0 + h) − g(0)
h2 − 0
= lim+
= 0.
h→0
h
h
2
Similarly, if we calculate the left-hand limit (where we only consider h < 0, denoted
limh→0− we get
−h2 − 0
g(0 + h) − g(0)
= lim−
= 0.
lim−
h→0
h→0
h
h
Since both the left and right-hand limits are equal to 0, it must be that the (two-sided)
limit is 0, and we get that g 0 (0) = 0. To summarize, we have that
(
−2x, if x < 0
g 0 (x) =
.
2x,
if x ≥ 0
So, g 0 (x) = 2|x|. Just as |x| was shown (in class) to be differentiable everywhere except
x = 0, the same argument shows that g 0 is differentiable everywhere except x = 0.
4. Below is the graph of a function f . Use its graph to make a rough sketch of the graph
of its derivative f 0 .
y
x
Solution: The graph of f 0 should look something like:
y
x
3
1
5. Find the equation(s) of the line(s) tangent to y = x−1
which pass through the point
(0, 1). Sketch and label the graph of the function as well as the tangent line(s).
Solution: A similar calculation to that in question 2 shows that, for x 6= 1, f 0 (x) =
1
− (x−1)
2 . Suppose the line tangent to f and going through (0, 1) is tangent at the point
,
(x0 , y0 ). Then we have two formulas for the slope of the tangent line: f 0 (x0 ) and xy00−1
−0
and so they have to be the same. That is,
1
y0 − 1
−
=
= f 0 (x0 ) =
2
(x0 − 1)
x0 − 0
1
x0 −1
−1
x0
If we try to solve this expression for x0 we get
x20 − 4x0 + 2 = 0,
which has solutions
√
√
16 − 4 · 2
x0 =
= 2 ± 2.
2
So, the equations of the tangent lines are given by
4±
y−1=−
1
√
(x − 0),
(2 ± 2 − 1)2
or
y =1−
1
√ x.
3±2 2
The curves are shown below.
2 y
1
x
−1
1
−1
−2
−3
4
2
3
4
.