MATHEMATICS 120 PROBLEM SET 4 – SOLUTIONS − 1. Let f and g

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MATHEMATICS 120 PROBLEM SET 4 – SOLUTIONS
sin x
and g(x) = − cot x. Find all values of x at which f 0 (x) and g 0 (x)
1 + cos x
exist and are equal.
1. Let f (x) =
We have
f 0 (x) =
cos x(1 + cos x) − sin x · (− sin x)
cos x + 1
1
=
=
(1 + cos x)2
(1 + cos x)2
1 + cos x
1
1
(we used that sin2 x + cos2 x = 1). Thus f 0 (x) = g 0 (x)
=
2
1 − cos2 x
sin x
1
1
only if
=
. Hence 1 + cos x = 1 − cos2 x, so that cos x = 0 or −1. But
1 + cos x
1 − cos2 x
if cos x = −1, then f (x) and g(x) are not defined. Therefore we must have cos x = 0, so
that x = π/2 + nπ, n = 0, ±1, ±2, . . .. At these values of x we have f 0 (x) = g 0 (x) = 1.
and g 0 (x) =
2. Find all values of a such that the graph of y = sin(ax) cos(ax) has a horizontal tangent
at the point where x = π.
Since y = sin(ax) cos(ax) = 12 sin(2ax), we have y 0 = a cos(2ax). For a horizontal tangent
at x = π we need either a = 0 or cos(2aπ) = 0. The second equation holds if and only
if 2aπ = π/2 + nπ, n = 0, ±1, ±2, . . .; equivalently, a = 1/4 + n/2, with n as above. The
answer is: a = 0, or a = 1/4 + n/2, n = 0, ±1, ±2, . . ..
3. Let f (x) = x2 + 4x + 10 cos x. Prove that there is a point a in (π/2, 10π) such that the
line tangent to the graph y = f (x) at the point (a, f (a)) goes through the origin. (Hint:
Find the equation of the tangent line for general a. When does this line go through the
origin? Does this equation have a solution in the given interval?)
The equation of the tangent line at x = a is y = f (a) + f 0 (a) · (x − a). Plugging in the
expression for f (a) and f 0 (a) = 2a + 4 − 10 sin a, we get
y = a2 + 4a + 10 cos a + (2a + 4 − 10 sin a)(x − a).
This line goes through (0, 0) if and only if
a2 + 4a + 10 cos a − a(2a + 4 − 10 sin a) = 0,
or equivalently a2 − 10 cos a − 10a sin a = 0. We need to prove that this equation has a
solution in (π/2, 10/pi). Let g(a) = a2 − 10 cos a − 10a sin a, then g(a) is a continuous
function of a, g(π/2) = π 2 /4 − 5π < 0, g(10π) = 100π 2 − 10 cos(10π) > 100π 2 − 10 > 0.
By the Intermediate Value Theorem, we must have g(a) = 0 for some a between π/2 and
10π.
1
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