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evablogger [386]
4 years ago
15

Consider the error in using the approximation e−x≈1−x on the interval [−1,1]. (a) reasoning informally, on what interval is this

approximation an overestimate? 1 an underestimate? (for each, give your answer as an interval or list of intervals,
e.g., to specify the intervals −0.25≤x<0.5 and 0.75
Mathematics
1 answer:
crimeas [40]4 years ago
6 0

e^{-x} \approx 1 - x

is a classic approximation, true for small x.

The next term in the polynomial expansion will be +kx^2 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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Solve each equation using the quadratic formula. Find the exact solution, then approximate the solution to the nearest hundredth
Anika [276]

Answer:

x_1=\frac{5+\sqrt{33}}{4}≈2.69

x_2=\frac{5-\sqrt{33}}{4}≈-0.19



Step-by-step explanation:

 To solve this problem you  must apply the proccedure shown below:

1. You have that the quadratic formula is:

x=\frac{-b+/-\sqrt{b^{2}-4ac}}{2a}

2. To solve the quadratic equation you must substitute the values. So, you have that:

Rewrite the equation:

2x^{2}-5x-1=0

a=2\\b=-5\\c=-1

Then you have:

 x=\frac{-(-5)+/-\sqrt{(-5)^{2}-4(2)(-1)}}{2(2)}

3. Therefore, you obtain the following result:

x_1=\frac{5+\sqrt{33}}{4}≈2.69

x_2=\frac{5-\sqrt{33}}{4}≈-0.19


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