Math 201
Exam 2 Key
1-2. If
(a) Find f 0 (x).
Solution.
(b) Find f 00 (x).
Solution.
Section 3
Spring 2013
f (x) = x4 + 8x3 + 200.
f 0 (x) = 4x3 + 24x2 .
f 00 (x) = 12x2 + 48x.
(c) Find the intervals of decrease and increase.
Solution. We need critical points. In this case, c such that f 0 (c) = 0. Need
f 0 (c) = 0 ⇐⇒ 4c3 +24c2 = 0 ⇐⇒ c3 +6c2 = 0 ⇐⇒ c2 (c+6) = 0 ⇐⇒ c = −6, 0.
Thus the critical points are -6 and 0. Let’s test the sign of f 0 on the intervals
(−∞, −6), (−6, 0) and (0, ∞). Since −7 ∈ (−∞, −6) and f 0 (−7) = −196 < 0,
it follows that f 0 < 0 on (−∞, −6) which implies that f decreases on
(−∞, −6). Since −1 ∈ (−6, 0) and f 0 (−1) = 20 > 0, it follows that f 0 > 0 on
(−6, 0) which implies that f increases on (−6, 0). Finally, since 1 ∈ (0, ∞)
and f 0 (1) = 28 > 0, it follows that f 0 > 0 on (0, ∞) which implies that f
increases on (0, ∞).
(d) Find the local maximum and minimum values.
By the First Derivative Test, since f 0 switches from negative to positive at
the critical value x = −6, (−6, f (−6)) = (−6, −232), this is a minimal point.
Since f 0 does not switch signs at the critical value x = 0, the point
(0, f (0)) = (0, 200) is neither maximal nor minimal.
(e) Find the intervals of concavity and inflection points.
Solution. Need the sign of f 00 (x). f 00 (x) = 0 ⇐⇒ 12x2 + 48x = 0 ⇐⇒
x2 + 4x = 0 ⇐⇒ x(x + 4) = 0 ⇐⇒ x = −4, 0. Let’s test the sign of f 00 on the
intervals (−∞, −4), (−4, 0) and (0, ∞). Since −5 ∈ (−∞, −4) and
f 00 (−5) = 60 > 0, it follows that f 00 > 0 on (−∞, −4) which implies that f is
concave up on(−∞, −4). Since −1 ∈ (−4, 0) and f 00 (−1) = −36 < 0, it
follows that f 00 < 0 on (−4, 0) which implies that f is concave down on(−4, 0).
Finally, since 1 ∈ (0, ∞) and f 00 (1) = 60 > 0, it follows that f 00 > 0 on (0, ∞)
which implies that f is concave up on(0, ∞). Since concavity changes at
(−4, f (−4)) = (−4, −56) and (0, f (0)) = (0, 200), these are inflection points.
(f) Sketch the graph of f .
Solution.
∅
200
100
–8
–6
–4
–2
2
x
–100
–200
3. Find the equation of the tangent line to
y=
2x
x+1
at (1, 1).
Solution. We need the slope m which will be given by the derivative of y evaluated
0 −(x+1)0 (2x)
2x
2
at (1, 1). y = x+1
=⇒ y 0 = (x+1)(2x)
= (x+1)(2)−(1)(2x)
= 2x+2−2x
= (x+1)
2.
(x+1)2
(x+1)2
(x+1)2
2
2
1
This evaluated at (1, 1) yields m = (1+1)2 = 4 = 2 . Hence, the equation of the
2x
tangent line to y = x+1
at (1, 1) is then given by
1
y−1=
(x − 1).
2
4. (a) Suppose that F (x) = f (g(x)), g(3) = 6, g 0 (3) = 4, f 0 (3) = 2 and f 0 (6) = 7.
Find F 0 (3).
Solution. F 0 (3) = (f (g(3)))0 = f 0 (g(3))g 0 (3) = f 0 (6)(4) = (7)(4) = 28.
√
(b) Find the derivative√of f (x) = e x .
u 0
Solution. Set u := x. We want (eu )0 . We know this is equal
to e
u . Thus
√
√ √
√
√
1
0
x 0
u 0
u 0
x
0
x 1/2 0
x
1/2−1
f (x) = (e ) = (e ) = e u = e ( x) = e (x ) = e
x
=
2
√
e x
√ .
2 x
√
(c) Find the derivative of f (x) = cos( x). √
Solution. cos u = − sin u · u0 . Setting u = x = x1/2 , we get u0 = (1/2)x−1/2 ,
thus
√
f 0 (x) = (cos( x))0
√
= − sin( x) · (x1/2 )0
√
= − sin( x) · (1/2)x−1/2
5. Use logarithmic differentiation to find the derivative of
y = (cos x)x .
Solution. y = (cos x)x =⇒ ln y = ln(cos x)x = x ln(cos x). Differentiating implicitly
we obtain then
y0
= (x)0 (ln(cos x)) + x(ln(cos x))0
y
(cos x)0
= (1)(ln(cos x)) + x
cos x
− sin x
= ln(cos x) + x
cos x
= ln(cos x) − x tan x
which implies
y 0 = y (ln(cos x) − x tan x)
= (cos x)x (ln(cos x) − x tan x) .
6. (a) Find the linearization of f (x) = cos x at a = π2 .
0 π
Solution. f (x) = cos x =⇒ f 0 (x) = − sin x. Therefore
f
2
= −1. Thus the
π
π
π
linearization (tangent line to the curve at 2 , f 2 = 2 , 0 is given by
π
.
y − 0 = −1 x −
2
(b) Use your answer in (a) to obtain an approximation to cos 3π
.
7
3π
Solution. Set x = 7 in the linearization of f (x) = cos x at a =
above:
3π π
(6 − 7)π
π
y=−
−
=−
= .
7
2
14
14
7. Find the equation of the tangent line to the curve
π
2
obtained
xy + 2x + 3x2 = 4
at the point (1, −1).
Solution. We have the point (1, −1), we need the slope m which is y 0 at (1, −1).
We use implicit differentiation to obtain y 0 : (xy + 2x + 3x2 )0 = 40 ⇐⇒
y + 2 + 6x
(xy)0 + (2x)0 + (3x2 )0 = 0 ⇐⇒ y + xy 0 + 2 + 6x = 0 =⇒ y 0 = −
. If x = 1
x
(−1) + 2 + 6(1)
and y = −1, then y 0 = −
= −7, thus the slope m = −7, hence the
1
tangent line has y − (−1) = −7(x − 1) or y = −7x + 6 as an equation.
8. If 1200 cm2 of material is available to make a box with a square base and an open
top, find the largest possible volume of the box.
Solution. If the box has length `, width w and height h, then the volume of the
box is given by
V = `wh.
Because the box has a square base, ` = w and therefore
V = w2 h.
This formula depends on two variables and we must get rid of one. Recall that the
area of the base is w2 . The area of each wall is wh and there are 4 such walls.
Finally, remember there is no top. Thus, the area of the box, which should be
1200 cm2 , is given by
1200 = A = w2 + 4wh
and therefore
1200 − w2
.
4w
Substituting in the formula for the volume, we obtain
1200 − w2
1200 − w2
2
=w
,
V =w
4w
4
h=
or
V = 300w −
w3
.
4
To find critical values we find
V 0 = 300 −
3w2
,
4
end equalizing to zero, we obtain
2
300 − 3w4 = 0 =⇒ 3w2 = 1200 =⇒ w2 = 400 =⇒ w = 20 cm (since w ≥ 0). By the
nature of the problem, this value of w yields the maximum volume, which is:
(20)3
V = 300(20) −
= 4000cm3 .
4