Calculus Of One Real Variable – By Pheng Kim Ving

8.3 
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1. TangentLine Approximations 
In general, if f(a + h) is difficult to determine while f(a) is readily determined and h is small, we may have to
approximate f(a + h) by a process that uses f(a).
In the 1st example above, h
= 0.2, and in the 2nd, h
= – 0.7. We'll
show that we can approximate the value of a function f at a point a
+ h by using
the tangent line to the graph of f
at
point a if h is small or a
+ h is near a.

Let f be
a differentiable function on an interval containing a
and x and h
= x – a so that x = a
+ h. See Fig.
1.1. Let (t,
y) be an arbitrary point on
the tangent line at a
other than the point (a,
f(a)). The equation of this tangent line is ( y –
f(a))/(t
– a) = f '(a), or:
y = f(a) + f '(a)(t – a).
So the height (or depth) at x = a + h
of the tangent line is f(a) +
f '(a)(x – a) = f(a) + h f '(a). Intuitively we can
approximate the height (or depth) f(a + h) at a
+ h of the
graph of f by the height
(or depth) f(a) + h f '(a)
at a + h of the tangent line at a. That is, at a + h the actual value of f is f(a + h) and its approximate value is f(a) + h
f '(a).
We
now prove that we can do that approximation. Let E(h)
be the difference between the actual and approximate values of f
at a + h, so that E(h) = f(a
+ h)
– ( f(a) + h f '(a))
= f(a + h) – f(a) – h f '(a).
Thus:
f(a + h) = f(a) + h f '(a) + E(h).
We want to compare E(h) and h when h is small or a + h is near a. We have:
Consequently, when h is sufficiently small, E(h) is very small. In fact, E(h) approaches 0 faster than h does. Hence
we can approximate f(a + h) by f(a) + h f '(a)
for small h:
As approximation [1.1] approximates the height (or depth) of a curve by that of a tangent line, it's called tangentline approximation or linear approximation. 
Recall that h may be positive or negative. For example:
If we utilize differential notation with dx = x – a = h then we obtain:

Example 1.1
iii. To approximate f(x), select a
that's closest to x
such that f(a) is readily determined,
write f(x) = f(a
+ h) where h =
x
– a, and
approximate f(a + h), as done in the above example.
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2. Error Of Approximation 
The difference:
E(h) = f(a + h) – ( f(a) + h f '(a))
between the actual value f(a + h) and approximate value f(a)
+ h f '(a)
of f
at a + h as given by the equation f(a + h)
=
f(a) + h f '(a) +
E(h) is the error of the approximation.
If f ''(x)
> 0 over an interval containing both a
and a + h, then the graph of f
is concave up over that interval, so it lies
above the tangent line at a,
thus the approximate value is smaller than the actual value, as a consequence
the error E(h)
is positive. Refer to Fig. 1.1. Similarly, if f ''(x)
< 0 over an interval containing both a
and a + h, then the error E(h) is
negative.
To see what may happen if f ''(x)
changes sign or the graph of f
changes concavity when x
goes from a to a + h, let's
look at an example shown in Fig. 2.1. Here f ''(x)
> 0 for b < x < p, f ''(x)
< 0 for p < x < c, a
and a + h_{1} are where
f ''(x)
> 0, and a + h_{2} and a + h_{3} are where f ''(x)
< 0. Clearly E(h_{1}) > 0, E(h_{2}) > 0, but E(h_{3}) < 0. Hence if f ''(x)
changes sign or the graph of f
changes concavity when x
goes from a to a + h, there's no conclusion about the sign of
the error; it may be positive ( like E(h_{2})) as well as negative ( like E(h_{3})). It's important that either f ''(x)
> 0 or f ''(x)
<
0 over an interval containing both a
and a + h to make sure that either E(h) > 0 or E(h) < 0 respectively.
Fig. 2.1 Sign of f
''(x) must be same over 
We now set out to find a bound
or (upper) limit on the magnitude (size, absolute value) of the error E(h). Refer to Fig.
1.1. If we move the point a
+ h by
changing the value of h
then clearly E(h) may or may not change. So E(h) and thus a
bound on E(h) depends on the displacement h. Now look at Fig. 2.2. The functions f_{1} and f_{2} have the same value f_{1}(a)
= f_{2}(a) and the same derivative f_{1}'(a)
= f_{2}'(a)
at a, and thus
the same tangent line at a.
The approximations:
The slopes of the tangent lines
are the same at a,
ie f_{1}'(a)
= f_{2}'(a).
Then both increase as we go from a
to x, and f_{2}'
increases faster than f_{1}' does, ie f_{2}'' > f_{1}''. So the difference between the curvatures is caused by the
difference between
Fig. 2.2 E(h) also depends on the 2nd derivative. 
the 2nd derivatives.
Consequently E(h) and hence a bound on E(h) also depends on the
2nd derivative f '' on [a,
a + h]
if h > 0 or
[a + h, a] if h
< 0. Note that the larger the 2nd derivative on [a, a
+ h], the
larger the curvature on
[a, a + h] and the larger the error at a + h.
The following theorem can be used to determine a bound on
the magnitude of the error E(h). It shows that E(h) and a
bound on its magnitude E(h) depend on both h
and f '' on [a,
a + h]
if h > 0 or
[a + h, a] if h
< 0.
Let f
be a twice differentiable function on an interval containing a and a + h, and let E(h) be the error of the f(a + h) = f(a) + h f '(a) + E(h). Then there exists b between a and a + h such that:

Proof
Suppose h > 0.
Let x = a + h, so that h
= x – a and E(h)
= E(x – a). For any t
in dom( f ) let h_{2} = t
– a, so that
E(h_{2}) = E(t – a). Let:
The proof is similar in the case where h < 0.
EOP
Interval Having Approximate Value At One End And Containing Actual
Value
Suppose, after finding an approximation and a bound on the
magnitude of the error, we're to obtain an interval having
the approximate value at one end and containing the actual value.
Suppose (approximate value) < (actual value). See Fig. 2.3. Then:
(actual value) – (approximate
value) < (bound on magnitude of error),
(actual value) < (approximate value) + (bound on magnitude of error),
(approximate value) < (actual value) < (approximate value) + (bound on
magnitude of error).
So the required interval is
[(approximate value), (approximate value) + (bound on magnitude of error)]. In
this case the
approximate value is at the lower end of the interval.
Fig. 2.3 If (approximate value) < 
Fig. 2.4 If (approximate value) > 
Now suppose (approximate value) > (actual value). See Fig. 2.4. Then:
(approximate value) – (actual value) < (bound on magnitude of error),
(approximate value) – (bound on magnitude of error) < (actual value),
(approximate value) – (bound on magnitude of error) < (actual value) < (approximate value).
So the required interval is [(approximate value) – (bound on
magnitude of error), (approximate value)]. In this case the
approximate value is at the upper end of the interval.
a. Utilize a suitable tangentline approximation to
determine an approximate value.
b. Determine the sign of the error of the approximation.
c. Get a bound on the magnitude of the error.
d. Obtain an interval having the approximate value at one end and
containing the actual value.
EOS
A bound on E(h) is (1/2)h^{2}M, not M,
which is a bound on  f ''(t).
We use a calculator to find an approximate value,
0.000957, of the quotient 0.49/512, which is ok because in this problem of
approximating square roots calculator is
implicitly banned only in finding square roots (of nonperfectsquares). Now
0.000957 is rounded up to 0.00096, which is
then used as a bound on E(– 0.7). What if the number is such that it would
be rounded down in ordinary
circumstances, for example 0.000953? Well, it wouldn't be rounded down to
0.00095, as 0.00095 < 0.000953 and thus
may or may not be a correct bound. We would still round 0.000953 up to 0.00096
to make sure we get a correct bound.
When you get a decimal value of a bound that's itself an approximate value
you'll have to round it up to make sure you
get a correct bound. If you're not sure whether the decimal value you obtain
from a calculator is approximate or exact
then just round it up.
Problems & Solutions 
a. Use tangentline
approximation to find an approximate value.
b. Determine the sign of the error.
c. Get a bound on the magnitude of the error.
d. Obtain an interval having the approximate value at one end and
containing the actual value.
Solution
2. For:
a. Utilize a suitable tangentline approximation to
determine an approximate value.
b. Determine the sign of the error.
c. Get a bound on the magnitude of the error.
d. Obtain an interval having the approximate value at one end and
containing the actual value.
a. Let f(x) = 1/x. So f '(x) = –1/x^{2}. Then:
A bound on the magnitude of the error is 0.000001125.
d. As 0.49925 <
1/2.003 and a bound on the magnitude of the error is 0.000001125, the desired interval
is [0.49925,
0.49925 + 0.000001125] or [0.44925,
0.499251125].
3. For e^{–}^{ 0.05}:
a. Employ tangentline approximation to find an approximate value.
b. Determine the sign of the error.
c. Get a bound on the magnitude of the error.
d. Obtain an interval having the approximate value at one end and
containing the actual value.
Solution
a. Let f(x) = e^{x}. Then f '(x) = e^{x}. We have:
4. For cos 62^{o}:
a. Use tangentline
approximation to find an approximate value.
b. Determine the sign of the error.
c. Get a bound on the magnitude of the error.
d. Obtain an interval having the approximate value at one end and containing
the actual value.
A bound on the magnitude of the error is 0.00031.
d. Since 0.46977 > cos 62^{o} and a bound on the magnitude of the
error is 0.00031, the required interval is [0.46977 –
0.00031, 0.46977] or [0.46946,
0.46977].
Note
Recall that when using
derivatives of trigonometric functions (here calculating the approximate value)
or things that are
obtained by utilizing them (here computing a bound on the magnitude of the
error) the angles must be in radians. That's
because derivatives of trigonometric functions are obtained by having angles be
in radians.
5. Prove that if a function f is three times differentiable
on an interval containing a
and x then
there exists b between a
and x such that:
Note: We can think of x as a + h so that x = a + h and x – a = h.
Solution
Suppose a < x. Let's define functions E and g as follows:
A similar argument leads to the same conclusion in the case where a > x.
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