Math 100 - Homework Set 3 (Taylor Approximation-Critical Points)
due date : Thursday March the 31st in class only
Basic skills required to work through the homework problems:
• computing and evaluating derivatives;
• determining an upper bound on the absolute value of a function on an interval;
• performing computations using powers, exponentials, trigonometric functions,
logarithms;
• computing and evaluating derivatives.
Learning Goals: After completing this problem set, you should be able to:
• find the Taylor polynomial of a specified degree for a function at a given point;
• use Taylor polynomials to approximate a function;
• use the remainder in Taylor’s formula to estimate the error in approximating a
function with its Taylor polynomial;
• find the critical points of a function on a given interval;
• find local and global maximum and minimum values of a function;
• use the Mean Value Theorem to show that under certain conditions the derivative
of a function takes a specific value at some point within a given interval, and apply
the theorem to prove a given mathematical relationship.
Taylor’s Formula with Remainder
Given the n−th degree Taylor polynomial of f (x) about x = a, if |f (n+1) (t)| ≤ M
for all values of t in the interval between a and x, then the remainder Rn (x) in Taylor’s
formula satisfies the inequality
|Rn (x)| ≤
M
|x
(n+1)!
− a|n+1
This means that when you use the Taylor polynomial of degree n to approximate
f (x), the error in the approximation can be estimated by finding an upper bound on
the (n + 1)-th derivative of f that works for all values in the intervals between x and
a, where a is the point about which you define the Taylor polynomial.
The Mean Value Theorem Let f be a function that satisfies the following hypotheses:
1. f is continuous on the closed interval [a, b].
2. f is differentiable on the open interval (a, b).
Then there is a number c in (a, b) such that
f 0 (c) =
f (b) − f (a)
b−a
1. (Warm-up Problems)
(a) Let f (x) be a function. What is the equation of the first degree Taylor polynomial of f (x) about x = a? How does this compare to the equation of the
tangent line to f (x) at x = a?
Solution: The first degree Taylor polynomial of f (x) about x = a is given
by
T1 (x) = f (a) + f 0 (a)(x − a)
The tangent line to f (x) at x = a is given by y − f (a) = f 0 (a)(x − a),
which rearranges to y = f (a) + f 0 (a)(x − a). Thus the first degree Taylor
polynomial and the tangent line are the same function.
(b) Suppose you wanted to estimate ln 0.5 using the first degree Taylor polynomial of f (x) = ln(x) about x = 1. Would your estimate be too high or too
low? (Hint: sketch a graph containing both f (x) and the first degree Taylor
polynomial).
Solution: If we sketch a graph of y = ln(x), then by part a) the first
degree Taylor approximation is just the tangent line to the curve at x = 1.
At x = 0.5 this line is above ln(0.5), and so your estimate would be too
high.
(c) Find an upper bound on the absolute value of f (x) = 3/x4 on the interval
[−2, −1].
Solution: On the interval [−2, −1], f (x) is positive and increasing. If
you sketch the graph of a function that is positive and increasing on an
interval, you will see that the largest absolute value of the function occurs
at the right endpoint.
Thus
the absolute value of the function at the right
3 endpoint, which is (−1)4 = 3 is an upper bound on the absolute value of
f (x) on the interval [−2, −1].
(d) Find an upper bound on the absolute value of f (x) =
[5, 10].
2x−1
7x2 −5
on the interval
Solution: On the interval [5, 10], the absolute value of the numerator of
f (x) is at most 19, and the absolute value of the denominator is always at
least 170. Thus |f (x)| ≤ 19/170 on the interval [5, 10].
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(e) Compute the 5th-degree Maclaurin polynomial for ex and use it to approximate
e.
Solution: The Maclaurin polynomial is just the Taylor polynomial about
x = 0. If f (x) = ex then
f (x) = f 0 (x) = ... = f (5) (x) = ex
and so f (0) = f 0 (0) = ... = f (5) (0) = e0 = 1. Thus the fifth degree
Maclaurin polynomial for y = ex is
T5 (x) = 1 + 1(x − 0) +
1
1
1
1
(x − 0)2 + (x − 0)3 + (x − 0)4 + (x − 0)5
2!
3!
4!
5!
We are interested in estimating e = f (1).
e = f (1) ≈ T5 (1) = 1 + 1 +
1 1
1
1
+ +
+
2 6 24 120
2. (a) Using a suitable
linear approximation of f (x) =
√
imation 1.1 ≈ 1.05 is reasonable.
√
(= 2.716̄)
x, explain why the approx-
√
Solution: f 0 (x) = 1/(2 x) so f 0 (1) = 1/2 and the linear approximation
about x = 1 is L(x) = 1 + (1/2)(x − 1). Substituting x = 1.1 gives
L(1.1) = 1.05.
√
(b) Using the linear approximation of the function
f
(x)
=
1 − x about x = 0
√
1.1.
Compare
your
result with the
(i.e. a = 0), approximate
the
number
√
approximation of 1.1 given in part (a) and discuss.
√
Solution: Now f 0 (x) = −1/(2 1 − x) and f 0 (0) = −1/2 so L(x) = 1 −
1/2(x) and L(−0.1)
√ = 1.05. This L(−0.1) is an appropriate approximation
for f (−0.1) = 1.1. We get the same final approximation as in (a) since
we are just translating the original function.
(c) Can you use √the linear approximation of f (x) =
approximate 1.1? Why?
√
1 − x about x = 1 to
Solution: No, since f 0 (1) does not exist.
3. Consider the function f (x) = ex cos(x).
Calculate Tn (x), the n-th degree Taylor polynomial to f (x) about x = 0, for the
cases:
(i) n = 0 (the constant approximation)
(ii) n = 1 (the linear approximation)
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(iii) n = 2 (the quadratic approximation)
(iv) n = 3 (the cubic approximation)
Solution: We begin by computing appropriate derivatives.
f (0) = 1
f 0 (x) = ex cos(x) + ex × (− sin(x)) = ex (cos(x) − sin(x))
f 0 (0) = 1
f 00 (x) = ex (cos(x) − sin(x)) + ex (− sin(x) − cos(x)) = −2ex sin(x)
f 00 (0) = 0
f 000 (x) = −2ex sin(x) − 2ex cos(x) = −2ex (sin(x) + cos(x))
f 000 (0) = −2
Thus the desired polynomials are
T0 (x) = 1,
T1 (x) = 1 + x,
T2 (x) = 1 + x + 0x2 = 1 + x,
T3 (x) = 1 + x + 0x2 +
−2
(x
3!
− 0)2 = 1 + x − 31 x3 .
4. (a) Find the 3rd-degree Taylor polynomial to y = (1 + x)3/2 at x = 3.
Solution: We have f (x) = (1 + x)3/2 . Thus
3
· (1 + x)1/2
2
3
f 00 (x) =
(1 + x)−1/2
4
−3
f (3) (x) =
(1 + x)−3/2
8
f 0 (x) =
In particular, f (3) = 8, f 0 (3) = 3, f 00 (3) = 3/8 and f (3) (3) = −3/64. Thus
1 3
1 3
· (x − 3)2 − · (x − 3)3
2 8
6 64
3
1
= 8 + 3(x − 3) + (x − 3)2 −
(x − 3)3
16
128
T2 (x) = 8 + 3(x − 3) +
(b) Use your result from part a) to estimate 4.13/2 .
Solution: 4.13/2 = f (3.1) ≈ T3 (3.1) = 8 + 3 ·
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1
10
+
3
16
·
1
100
−
1
128
·
1
.
1000
(c) Give an upper bound on the magnitude of the error of your estimate from b).
Solution: To apply Taylor’s formula with remainder, we have to find a
number M such that |f (4) (x)| ≤ M for all x in the interval [3, 3.1] (Remember, our a-value is 3, and the x-value we’re interested in is 3.1). If we take
9
the derivative of f (3) (x) from part a), we find that f (4) (x) = 16
(1 + x)−5/2 .
On the interval [3, 3.1], f (4) (x) is positive and decreasing, since the denominator gets bigger as x gets bigger. Thus, its largest absolute value will
occur at the left endpoint (draw a picture of a positive, decreasing function
9
9
to see why). So we can take M = |f (4) (3)| = 16·2
5 = 512 .
By Taylor’s formula the error in our approximation, which is R3 (3.1), satisfies:
|R3 (3.1)| ≤
1
9
9
· · |3.1 − 3|4 =
512 4!
512 · 24 · 10000
(≈ 0.00000007)
Thus an upper bound on the magnitude of the error in the estimate from
9
part b) is 512·24·10000
.
5. (a) Approximate
2.
√
101 by using appropriate Taylor polynomials of degree 1 and
√
Solution: With f (x) = x we have f 0 (x) = (1/2)x−1/2 and f 00 (x) =
(−1/4)x−3/2 , so f (100) = 10, f 0 (100) = 1/20, and f 00 (100) = −1/4000.
Using the approximations about x = 100, we have P1 (x) = 10 + (1/20)(x −
100) (this is just the linear approximation) and P2 (x) = 10 + (1/20)(x −
100) − (1/4000)(x − 100)2 . Substituting x = 101 gives the respective
approximate
values 10 + 1/20 = 10.05 and 10 + 1/20 − 1/4000 = 10.04975
√
for 101.
(b) Use Taylor’s formula with remainder to
√ estimate the accuracy of the approximations of degrees 1 and 2 of f (x) = x, for 95 ≤ x ≤ 105.
Solution: Taylor’s formula with remainder says
|Rn (x)| ≤
M
|x − a|n+1 ,
(n + 1)!
where |f (n+1) (t)| ≤ M for all t between x and a. The key is finding a
suitable value for M .
For degree 1, we have |f 00 (x)| = 14 x−3/2 . Now |f 00 (x)| < 14 · 81−3/2 = 4·91 3 for
95 ≤ x ≤ 105 (this is just an estimate, we substitute 81 to get an upper
bound on f 00 since f 00 is a decreasing function and 81 is a perfect square).
So, we take M to be this value, and then |R1 (x)| ≤ 4·913 ·2! |x − 100|2 ≤
1·52
52
25
= 8·9
3 = 1944 ≈ 0.013.
4·93 ·2
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For degree 2, we have f 000 (x) = 38 x−5/2 . Now f 000 (x) < 38 · 81−5/2 = 8·93 5 for
95 ≤ x ≤ 105 (this is just an estimate, we substitute 81 to get an upper
bound on f 000 since f 000 is a decreasing function and 81 is a perfect square).
So, we take M to be this value, and then |R2 (x)| ≤ 8·935 ·3! |x − 100|3 ≤
3·53
53
125
= 16·9
5 = 944784 ≈ 0.00013.
8·95 ·3!
6. Find the critical numbers of f (x) = x + cos(2x) on the interval [0, π/2].
Solution:
We compute f 0 (x) = 1 − 2 sin(2x) and so f 0 (x) is defined for all x ∈ [0, π/2].
Now f 0 (x) = 0 for 2 sin(2x) = 1 and so for sin(2x) = 12 and so for 2x = π/2−π/3
and 2x = π/2 + π/3 (for x ∈ [0, π]) and so for x = π/12, 5π/12.
7. Determine the x-coordinates of local maxima and minima, if any, of the following
function:
 x−1
x < 1;
 e ,
x2 + 3
f (x) =
, x ≥ 1.

2(x + 1)
Solution: First note that the two pieces match at x = 1 so the function is
continuous. Using the quotient rule, the derivative of (x2 + 3)/(2(x + 1)) equals
2(x + 3)(x − 1)/(4(x + 1)2 ). The only value ≥ 1 for which this equals 0 is x = 1.
The derivative of ex−1 is never 0. So, the only possible local max or min is at
x = 1. However, since the function is increasing both to the right and left of
x = 1 (look at the sign of the derivative) there is in fact no max or min.
8. Use the Intermediate Value Theorem and the Mean Value Theorem to show that
the equation 7x − 1 − sin x = 0 has exactly one real root.
Solution: Let f (x) = 7x − 1 − sin x, a continuous and differentiable function
for all real x. Since f (0) = 0 − 1 − 0 = −1 < 0 and f (1) = 7 − 1 − sin 1 > 0,
by the Intermediate Value Theorem there is a root in the interval (0, 1), so we
know f (x) = 0 has at least one real root. To show there is exactly one real
root, suppose a < b are two solutions of f (x) = 0. Then, by the Mean Value
Theorem applied to f (x) on the interval [a, b] there is a point c in the interval
(a, b) such that f 0 (c) = (f (b) − f (a))/(b − a) = (0 − 0)/(b − a) = 0. But
f 0 (x) = 7 − cos x, which cannot be 0 since cos x never takes the value 7. So, we
have a contradiction to the existence of two roots, and there is indeed exactly
one real root.
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9. Suppose that a function f is twice differentiable on [0, 4] and that f (1) = f (2) = 0
and f (3) = 1. Show that:
(i) f 0 (a) = 1/2 for some point a in (0, 4).
(ii) f 00 (b) > 1/2 for some point b in (0, 4).
Solution: The first thing to remember is that if a function is differentiable
then it is continuous (this follows from the limit definition of the derivative)
and so in this circumstance f and f 0 are continuous.
(1)
We note that f (3)−f
= 12 . We may apply the mean value theorem to f for
3−1
a = 1, b = 3 to deduce that there is an c ∈ (1, 3) (and hence a ∈ (0, 4)) with
(1)
= 12 .
f 0 (a) = f (3)−f
3−1
You can also deduce that there is some d ∈ (2, 3) with f 0 (d) = 1 and some
e ∈ (1, 2) with f 0 (e) = 0. We use these choices in answering the second part.
For the second part it is now sensible to imagine f 0 (x) is the function under
consideration with f 00 (x) being its derivative. To make this explicit in this
solution let g(x) = f 0 (x) so that g 0 (x) = f 00 (x). We have g(d) = 1 for some
2 < d < 3 and g(e) = 0 for some 1 < e < 2. Now apply the Mean Value
Theorem with function g and a = e and b = d. Then there is a value k with
e < k < d and
1
g(d) − g(e)
≥
f 00 (k) = g 0 (k) =
d−e
2
noting that g(d) − g(e) = 1 and d − e < 2 (in fact 0 < d − e < 2).
10. Find the absolute maximum and absolute minimum values of the function
p
f (x) = (1 − x) · x + 3/2
on the interval [−1, 1].
Solution:
p
We compute f 0 (x) = √ 1
· (1 − x) − x + 3/2. Thus f 0 (x) = 0 for √ 1
·
2 x+3/2
2 x+3/2
p
(1 − x) = x + 3/2, which simplifies to 1 − x = 2x + 3, so x = −2
, and this
3p
is the only critical number inside the interval. We have f (−2/3) = 5/6 · 35 =
p
p
√
125/54, f (−1) = 1/2 · 2 = 2, f (1) = 0. The largest of these is the first
value (since 125/54 > 2) and hence this is the absolute maximum. The smallest
value, which is 0, is the absolute minimum.
11. (For bonus points) Suppose f (x) is a function that is differentiable for all x. Let
g(x) be the new function defined by g(x) = f (x) + f (1 − x2 ). Prove that g 0 (c) = 0
for some positive real number c with 0 < c < 1.
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Solution: We want to apply the Mean Value Theorem to g but we don’t know
any function values. We hunt for some useful facts using the strange expression
for g. We compute g(0) = f (0) + f (1) and g(1) = f (1) + f (0). And so
g(0) = g(1) although we don’t know much else. But then applying Mean Value
Theorem to g with a = 0 and b = 1, we have that there is some c ∈ (0, 1) with
= 0. Now to be honest we haven’t verified that g is differentiable
g 0 (c) = g(1)−g(0)
1−0
(and hence continuous) so we compute g 0 (x) = f 0 (x) + f 0 (1 − x2 ) × (−2x) and
so g is differentiable for all x ∈ R.
12. (For bonus points) Use the Mean Value Theorem to show that sin x < x for all
x > 0.
Solution:
To use the Mean Value Theorem, we need a function. Let f (x) = x − sin(x).
This function is continuous (sum of continuous functions) and differentiable
(f 0 (x) = 1 − cos(x)). Now if sin(c) ≥ c for some 0 < c, we have that f (c) ≤ 0.
We already know f (0) = 0. If we apply the Mean Value Theorem to f with
a = 0 and b = c, we have that there is a value d with 0 < d < c with
f 0 (d) =
f (c) − f (0)
≤0
c−0
But f 0 (x) = 1 − cos(x) and cos(x) ≤ 1 for all x and so f 0 (x) ≥ 0 for all x and
so f 0 (d) ≥ 0. Hence we have a contradiction if cos(d) < 1 but we can have
f 0 (c) = 0 and cos(d) = 1 and have no apparent contradiction.
Let us repeat the argument by avoiding the possibility above with cos(d) = 1
for 0 < d < c. One way to do this is to note that f (c) > 0 for c > 1 since
sin(x) ≤ 1. So we may restrict our attention to values of c with 0 < c < 1. Now
with 0 < d < c < 1, we have that cos(d) < 1 (because cos(x) = 1 for x = 0 and
x = 2π but for no intermediate value of x) and so our contradiction above now
follows. Thus sin(x) < x for all x.
An alternate solution is to proceed as follows. Consider the function f (x) =
sin(x). If we have sin(x) ≤ 0, then sin(x) < x for all x > 0. Also for x > 1, the
inequality is trivial. So we may assume sin(x) > 0 and 0 < x < 1. Note that
f (0) = 0. Now using the Mean Value theorem with a = 0 and b = x, then for any
= sin(x)
.
given x > 0, there is a c with 0 < c < x and f 0 (c) = cos(c) = sin(x)−sin(0)
x−0
x
sin(x)
Now cos(c) ≤ 1 for 0 < c < 1 < 2π and so x < 1 and so sin(x) < x.
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