EXAM 2 SOLUTIONS
(1) (a)
(x3 + 3x2 )0 = 3x2 + 6x.
(b)
(3 tan−1 (x))0 =
3
1 + x2
(c)
(sin(x) + cos(x))0 = cos(x) − sin(x)
(d)
1
−1 1
( + ln(x))0 = 2 +
x
x
x
(2) (a)
1
1
1
sin( √
) = cos( √
)( √
)0
1 + x2
1 + x2
1 + x2
−x
1
)
= cos( √
2
1 + x (1 + x2 )3/2
(b)
(x2 esin(x) )0 = 2xesin(x) + x2 esin(x) cos(x)
(c)
(x(1 + x2 )500 )0
= (1 + x2 )500 + x500(1 + x2 )499 (2x)
= (1 + x2 )500 + 1000x2 (1 + x2 )499
(d)
p
1
1
1
( ln(x))0 = ln(x)−1/2 = p
2
x 2x ln(x)
(3) ??
(4) Let f (x) = x1/3 . Then f 0 (x) = 13 x−2/3 . The tangent line to the point (a, a1/3 ) is given by the equation,
y
f (a) + f 0 (a)(x − a)
1
= a1/3 + 2/3 (x − a).
3a
=
The idea of linear approximation is that the this line will closely approximate the graph of f (x) for x-values
near the value of a. We are interested in approximating (27.1)1/3 , and so we take a = 27 for example. Then
the above tangent line becomes,
1
y = 3 + (x − 27).
27
We are then interested in the value when x = 27.1 which gives,
y = 3+
1
811
=
270 270
(5) The line which passes through (1, 2) and is tangent to the curve x3 + xy2 + y3 = 13 is of the form,
y−2 =
dy
(x − 1)
dx
Date: April 14, 2008.
1
where
dy
dx
is the value of the derivative at x = 1. But to find
dy
dx
we must differentiate implicitly,
d
d 4
(x + xy2 + y3 ) =
(13)
dx
dx
d
d
4x3 + y2 + x (y2 ) + (y3 ) = 0
dx
dx
dy
dy
4x3 + y2 + x(2y) + 3y2
= 0
dx
dx
Plugging in x = 1 and y = 2 and solving for
dy
dx
we get,
dy
4(1)3 + (2)2
−1
=−
=
.
dx
3(2)2 + 1(2 ∗ 3)
2
Hence the equation of the tangent line is
1
y = 2 − (x − 1).
2
(6) The critical points of the differentiable function f (x) = x sin(x) + cos(x) are the points c which satisfy,
f 0 (x) = sin(x) + x cos(x) − sin(x) = x cos(x) = 0.
But x cos(x) = 0 when x = 0, ±π/2. Thus, we just need to compare the values of f (x) a the 3 critical points
and the 2 endpoints of the interval ±π at the points −π, π/2, 0, π/2, π and choose the biggest and smallest
values. We have,
f (−π) = (−π) sin(−π) + cos(−π) = (−π)0 − 1 = −1
f (−π/2) = (−π/2) sin(−π/2) + cos(−π/2) = (−π/2)(−1) + 0 = π/2
f (−0) = (0) sin(0) + cos(0) = (0) + 1 = 1
f (π/2) = (π/2) sin(π/2) + cos(π/2) = (π/2)(1) + 0 = π/2
f (π) = (π) sin(π) + cos(π) = π(0) − 1 = −1
Hence the absolute maximum value of π/2 occurs at x = ±π/2 and the absolute minimum value of −1
occurs at x = ±π.
(7) (a) The function f is differentiable, hence continuous on the interval [0, 3]. Therefore,by the extreme value
theorem, it has an absolute minimum on this interval. This minimum cannot occur at the endpoints since
we already have a point x = 2 which is contained in the interval (0, 3) and which satisfies f (2) < f (0)
and f (2) < f (3). So the absolute minimum occurs at a ∈ (0, 3) and so must actually be a local minimum
of the function f (x). But by Fermat’s theorem, a local minimum must be a critical point and since f (x)
is assumed to be differentiable, at the critical point we have f 0 (a) = 0.
(b) By the intermediate value theorem, for any value α, where
f (0) = 5 ≤ α ≤ f (3) = 6,
there exists a x = c ∈ [0, 3] such that f (c) = α. We choose α = 5.5 and this part follows.
(c) By way of contradiction, suppose that there are two points a, b ∈ [0, 3] such that
f (a) = f (b) = 5.5.
Let us suppose that a < b. Then by the mean value theorem, there is a point c ∈ (0, a) such that
f (a) − f (0) 5.5 − 5
=
> 0.
a−0
a
By Rolle’s theorem, there is a point d ∈ (a, b) such that
f 0 (c) =
f 0 (d) = 0.
But c < d and f 0 (c) > f 0 (d) = 0 which is a contradicts the fact that f 0 (x) was an increasing function.
Note: The fact that f (1) = 2 is completely irrelevant for this part of the problem.
2