Exam 2 Solution
4. Version 1 and 2
Let f (x) = xlnx
(A) Find the domain of f (1 point) and all zeros of f (1
point).
Logarithmic functions only accept positive values, therefore the domain of f is x > 0
Zeros of f exist where f = 0
xlnx = 0 therefore x = 0 or lnx = 0. However, x = 0 is
outside the domain. Therefore, the only zero is at x = 1,
because ln(1) = 0.
(B) Find f’(4 points). Determine where f is increasing
(1 point) / decreasing (1 point), or has local maxima /
minima (2 points).
Using the product rule, f 0 = (1)(lnx) + (x)( x1 ). Simplified that is f 0 = lnx + 1.
The function has horizontal tangents where the first derivative equals 0, is increasing where the the derivative is
positive and decreasing where the derivative is negative.
f 0 = lnx + 1 = 0 implies lnx = −1. Therefore, by the
definition of logarithms, x = e−1 .
Between 0 and e−1 , the function is decreasing.
We can
0
−2
use a test point in the interval to find f e
= ln e−2 +
1 = −2lne + 1 = −2 (1) + 1 = −1.
Between e−1 and ∞, the function is increasing. By test
point we see f 0 (e) = ln (e) + 1 = 1 + 1 = 2.
From this we see that the point (e−1 , −e−1 ) is a minimum.
1
(C) Find f” (2 points). Determine where f is concave
up / down (1 point) or has an inflection point (1 point).
f ” = x1 .
f ” 6= 0 for any x in the domain. Therefore, f has no
inflection points.
Test point f ”(1) = 11 = 1. Therefore, between 0 and ∞
f is convex or concave up. And never concave down.
(D) Find limx→0 f (x) (2 points)
We may rewrite the function as ln(x)
1 . In this case, we
x
may apply
Rule. Therefore, limx→0 f (x) =
1 L’Hopital’s
limx→ 0 −x1 . This simplifies to limx→ 0 (−x) = 0.
x2
(E) Sketch the graph of f (1 point), label intercepts (1
point), maxima / minima (1 point), and inflection points
(1 point).
4. Version 3 and 4
Let f (x) = lnx
x
(A) Find the domain of f (1 point) and all zeros of f (1
point).
2
Logarithmic functions only accept positive values, therefore the domain of f is x > 0
Zeros of f exist where f = 0
lnx
x = 0 therefore x 6= 0 and lnx = 0. However, x = 0 is
outside the domain. Therefore, the only zero is at x = 1,
because ln(1) = 0.
(B) Find f’(4 points). Determine where f is increasing
(1 point) / decreasing (1 point), or has local maxima /
minima (2 points).
( 1 )(x)−(1)(lnx)
Using the quotient rule, f 0 = x x2
. Simplified
1−ln(x)
that is f 0 = x2 .
The function has horizontal tangents where the first derivative equals 0, is increasing where the the derivative is
positive and decreasing where the derivative is negative.
f 0 = 1−ln(x)
= 0 implies lnx = 1. Therefore, by the defix2
nition of logarithms, x = e.
Between 0 and e, the function is increasing. We can use a
(1)
test point in the interval to find f 0 (1) = 1−ln
= 1−0
12
1 =
1.
Between e and ∞, the function is decreasing. By test
1−ln (e2 ) 1−2
point we see f 0 e2
= e2 = − e12 .
2
(e2 )
From this we see that the point (e, e−1 ) is a maximum.
(C) Find f” (2 points). Determine where f is concave
up / down (1 point) or has an inflection point (1 point).
f” =
f” =
(− x1 )x2 −2x(1−ln(x))
2
(x2 )
2 ln(x)−3
x3
=
2 ln(x)−3
.
x3
= 0 where lnx = 32 . Therefore, f has an
3
3
inflection point at x = e 2 .
Test point f ”(1) = 2 ln 1(1)−3
= − 31 = −3. Therefore,
3
3
3
between 0 and e 2 f is concave down and between e 2
and ∞ f is convex or concave up, based on test point
2 ln (e2 )−3
f e2 = (e2 )3 = 2(2)−3
= e−6 > 0.
e6
(D) Find limx→0 f (x) (2 points)
If we rewrite f(x) = ln (x) x1 then one sees that, as x
approach 0 from the positive side, x1 approaches ∞ and
ln (x) approaches −∞. Therefore, f approaches −∞.
1
lnx
Also, limx→∞ f (x) = 0 because limx→∞ x = lim x→ ∞ 1x =
0 according to L’Hopital’s Rule.
(E) Sketch the graph of f (1 point), label intercepts (1
point), maxima / minima (1 point), and inflection points
(1 point).
4