Answer Key for Exam 1, Math 105 (Section B)
1. If f(x) = ln (ex), then the chain rule says that
f0(x) = 1
ex
d
dx ex=1
exex= 1.
This is not surprising, since ln xand exare inverse functions of each other, which means that ln(ex) = x.
Hence f(x) = x, and therefore f0(x) = 1.
2. We have g(x) =
x2+ 6x+ 7 if x < 1
x2+ 7x+ 6 if 1 < x < 3
13x−3 if x≥3.
(a) g(x) is not continuous at x= 1, because it is not defined at x= 1. This is so even though lim
x→1−
g(x) =
12+ 6 ·1 +7 = 14 and lim
x→1+g(x) = 12+ 7 ·1 +6 = 14 also. Thus lim
x→1g(x) exists and equals 14, but there is
no g(1) for it to be equal to.
At x= 3 we have lim
x→3−
g(x) = 32+ 7 ·3 + 6 = 36 and lim
x→3+g(x) = 13 ·3−3 = 36, so lim
x→3g(x) exists and
equals 36, and also g(3) is defined; it equals 13 ·3−3 = 36. Therefore g(x) is continuous at x= 3 since the
limit equals the function value there.
(b)
g0(x) =
2x+ 6 if x < 1
2x+ 7 if 1 < x < 3
13 if x > 3.
It is pretty obvious that the first two pieces of this are never going to agree. As x→1−,g0(x)→2·1+ 6 = 8,
and as x→1+,g0(x)→2·1 + 7 = 9. This tells us that g0(1) does not exist, but actually we knew that
already—since there is no g(1), it is not possible that g0(1) could exist.
As x→3−,g0(x)→2·3 + 7 = 13, and for any x > 3 (so in particular as x→3+)g0(x) = 13, so g0(3) exists
and equals 13. We could therefore change 1 < x < 3 to 1 < x ≤3 and x > 3 to x≥3 in the expression for
g0(x), if we want.
(c)
g00(x) =
2 if x < 1
2 if 1 < x < 3
0 if x > 3.
Since g(1) and g0(1) do not exist, g00(1) does not exist, even though it equals 2 on either side of x= 1. g00 (3)
does not exist either since 2 6= 0.
4(a) If a(x) = x|x|, then the product rule says that
a0(x) = xd
dx |x|+|x|d
dx x=x|x|
x+|x| · 1 = |x|+|x|= 2|x|.
4(b) If b(x) = |ex|, then by the chain rule
b0(x) = |ex|
ex
d
dx ex=|ex|
exex=|ex|.
This is not too surprising since exis always positive; therefore |ex|=exfor all x. So in fact b(x) = ex, and
hence b0(x) = ex.
