Calculus Of One Real Variable By Pheng Kim Ving
Chapter 8: Applications Of The Derivative Part 2 Section 8.3: Tangent-Line Approximations

 

8.3
Tangent-Line Approximations

 

 

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1. Tangent-Line 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.

 

Fig. 1.1

 

 

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 tangent-line 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

 

 

Solution

EOS

 

Remarks 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.

 

Sign Of Error

 

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 + h1 are where
f ''(x) > 0, and a + h2 and a + h3 are where f ''(x) < 0. Clearly E(h1) > 0, E(h2) > 0, but E(h3) < 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(h2)) as well as negative ( like E(h3)). 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
an interval containing both a and
a + h to conclude about the sign
of the error.

 

Bound On Magnitude Of Error

 

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 f1 and f2 have the same value f1(a)
= f2(a) and the same derivative f1'(a) = f2'(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 f1'(a) = f2'(a). Then both increase as we go from a to x, and f2'
increases faster than f1' does, ie f2'' > f1''. 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.

 

Theorem 2.1

 

Let f be a twice differentiable function on an interval containing a and a + h, and let E(h) be the error of the
tangent-line approximation of f(a + h) using f(a), ie:

 

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 h2 = t a, so that
E(h2) = 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) <
(actual value) then approximate value is at lower
end of required interval.

 

Fig. 2.4

 

If (approximate value) >
(actual value) then
approximate value is at upper
end of required interval.

 

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.

Example 2.1

 

a. Utilize a suitable tangent-line 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.

 

Solution

EOS

 

 

A bound on |E(h)| is (1/2)h2M, 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 non-perfect-squares). 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.

 

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Problems & Solutions

 

a. Use tangent-line 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

 

 

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2. For:

 

 

a. Utilize a suitable tangent-line 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.

 

Solution

 

a. Let f(x) = 1/x. So f '(x) = 1/x2. 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].

 

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3. For e 0.05:
a. Employ tangent-line 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) = ex. Then f '(x) = ex. We have:

 

 

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4. For cos 62o:

a. Use tangent-line 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 bound on the magnitude of the error is 0.00031.

 

d. Since 0.46977 > cos 62o 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.

 

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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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