Download Solution to problems suggested in our last class Calculus I (Math 111)

Solution to problems suggested in our last class
Calculus I (Math 111)
• Any polynomial is continuous everywhere.
A simple consequence of the Direct Substitution Theorem. Indeed, for any polynomial f (x) and any
point a, we have
lim f (x) = f (a).
x→a
To prove the DST, use the limit laws. Start with a generic polynomial f (x) = c0 + c1 x + · · · + cn xn
and use the sum law to break limx→a f (x) into a sum of n + 1 terms. Use the constant multiple law
and the power law to handle each term of the sum.
• Interpreting limx→a f (x) = −∞.
Informally, we say that f (x) approaches −∞ as x approaches a. Quantitatively, it means that for
every number M (say, M = −101000 ) there is some δ > 0 such that for any x close to a, meaning 0 <
|x − a| < δ, we have f (x) < M .
Plot a graph of such a function x and draw a horizontal strip on the xy-plane corresponding to y ≤ M
and a vertical strip corresponding to |x − a| < δ. Try to see what the correspondence between the
quantitative version above and the picture you drew.
• Obtain the absolute maximum and minimum of f (t) = 2 cos t + sin(2t) on [0, π/2].
Clearly you must use the closed interval method. First notice that f is differentiable everywhere (in
particular, it is differentiable on the interval [0, π/2]).
First, let us plug in the extremes 0 and π/2 on f . Warning: many people have plugged those extreme
values on f 0 , DO NOT do this, this is not what the method asks for. We get f (0) = 2 and f (π/2) = 0.
Now we obtain the critical points for f , namely, the points where f 0 (t) = 0. To make our lives easier,
let us use the trig identity sin(2t) = 2 sin t cos t to simplify f , that is,
f (t) = 2 cos t(1 + sin t)
(1)
Deriving, we obtain
f 0 (t) = −2 sin t(1 + sin t) + 2 cos t(cos t) = −2 sin t + 2(cos2 t − sin2 t).
But by Pythagoras, sin2 t + cos2 t = 1, meaning cos2 t = 1 − sin2 t. Replacing cos2 t in the expression
above yields
f 0 (t) = 2(1 − sin t − 2 sin2 t) = 2(1 + sin t)(1 − 2 sin t).
Hence, the solutions to f 0 (t) = 0 occur either when sin t = −1 or when sin t = −1/2. We could
find t at this stage but it turns out that this is not needed. When sin t = −1 we know that t √
is not
2
2
in the range [0, π/2]. When sin t = 1/2, by Pythagoras,
cos
t
+
(1/2)
=
1,
hence
cos
t
=
±
3/2.
√
Since for t √∈ [0, π/2], cos t√≥ 0, we have cos t = 3/2 as the only possibility. In this case, one
has f (t) = 3(1 + 1/2) = 3 3/2.
1
By the closed interval
method, the absolute √
minimum is 0, attained at the extreme π/2. The absolute
√
maximum is 3 3/2, attained when cos t = 3/2 (when t = π/6 to be more explicit). See the graph
below.
2.5
2.0
1.5
1.0
0.5
0.5
1.0
1.5
2