Consider the error in using the approximation e−x≈1−x on the interval [−1,1]. (a) reasoning informally, on what interval is this
1 answer:
is a classic approximation, true for small x.
The next term in the polynomial expansion will be where k is a positive number. So our estimate 1-x is definitely an underestimate on both sides of x=0.
Since for negative xs the exponential rises exponentially and the line only linearly, the exponential exceeds the line for all negative x. For positive x, the line quickly goes negative while the exponential is always positive.
So, there's no interval for which our approximation is an overestimate.
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Answer:
Step-by-step explanation:
Answer:
x >= -7 ................(1a)
x >= 3 ...............(2a1)
Step-by-step explanation:
f(x) = .............(0)
find the domain.
To find the (real) domain, we need to ensure that each term remains a real number.
which means the following conditions must be met
x+7 >= 0 .....................(1)
AND
x^2+2x-15 >= 0 ..........(2)
To satisfy (1), x >= -7 .....................(1a) by transposition of (1)
To satisfy (2), we need first to find the roots of (2)
factor (2)
(x+5)(x-3) >= 0
This implis
(x+5) >= 0 AND (x-3) >= 0.....................(2a)
OR
(x+5) <= 0 AND (x-3) <= 0 ...................(2b)
(2a) is satisfied with x >= 3 ...............(2a1)
(2b) is satisfied with x <= -5 ................(2b1)
Combine the conditions (1a), (2a1), and (2b1),
x >= -7 ................(1a)
AND
(
x >= 3 ...............(2a1)
OR
x <= -5 ................(2b1)
)
which expands to
(1a) and (2a1) OR (1a) and (2b1)
( x >= -7 and x >= 3 ) OR ( x >= -7 and x <= -5 )
Simplifying, we have
x >= 3 OR ( -7 <= x <= -5 )
Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.
