Answer Key for Exam 1, Math 105 (Section D)
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, so f0(x) = 1.
2. We have g(x) =
x2−5x+ 6 if x≤1
4−x−x2if 1 < x ≤2
x2−3xif x > 2.
(a) lim
x→1−
g(x) = 12−5·1 + 6 = 2, and lim
x→1+g(x) = 4 −1−12= 2 also. Therefore lim
x→1g(x) exists and equals
2, and also g(1) is defined to be 12−5·1 + 6 = 2. Hence the limit equals the function value, so g(x) is
continuous at x= 1.
At x= 2 we have lim
x→2−
g(x) = 4 −2−22=−2 and lim
x→2+g(x) = 22−6 = −2. Thus lim
x→2f(x) exists and
equals −2, and also g(2) is defined to be 4 −2−22=−2. Therefore g(x) is continuous at x= 2 since the
limit equals the function value there.
(b)
g0(x) =
2x−5 if x < 1
−1−2xif 1 < x < 2
2x−3 if x > 2.
As x→1−,g0(x)→2·1−5 = −3, and as x→1+,g0(x)→ −1−2·1 = −3. Therefore g0(1) exists and
equals −3. We could then change x < 1 to x≤1 and 1 < x < 2 to 1 ≤x < 2 in the definition of g0(x), if we
want.
As x→2−,g0(x)→ −1−2·2 = −5, and as x→2+,g0(x)→2·2−3 = 1. Since these do not agree, g0(2)
does not exist.
(c)
g00(x) =
2 if x < 1
−2 if 1 < x < 2
2 if x > 2.
Since 2 6=−2, neither g00(1) nor g00 (2) can exist. g00(2) could not have existed anyway since g0(2) does not
exist.
4(a) If a(x) = |x|x−1, then the product rule says that
a0(x) = |x|d
dx x−1+x−1d
dx |x|=|x|¡−1x−2¢+x−1|x|
x=−|x|
x2+|x|
x2= 0.
4(b) If b(x) = ln |x|, then the chain rule says that
b0(x) = 1
|x|
|x|
x=1
x.
