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Elis [28]
3 years ago
11

How do I solve this?

Mathematics
1 answer:
vodomira [7]3 years ago
7 0
I’m assuming what you’re asking here is how to *factor* this expression. For that, let’s rearrange the expression into a more familiar form:

-c^2-4c+21

From here, we’ll factor out a -1 so that we have:

-(c^2+4c-21)

Let’s focus on the quadratic expression inside the parentheses. To find our factors (c + x)(c + y), we’ll need to find two terms x and y that multiply together to make -21 and add together to make 4. It turns out that the numbers -3 and 7 work out perfectly for that purpose (-3 x 7 = -21 and 7 + (-3) = 4), so substituting them in for x and y, we have:

(c + (-3))(c + 7)
(c - 3)(c + 7)

And adding back on the negative from a few steps earlier:

-(c - 3)(c + 7)
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Determine the values of xfor which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.(Enter
MrMuchimi

Answer:

The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

Step-by-step explanation:

Note: This question is not complete. The complete question is therefore provided before answering the question as follows:

Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0

The explanation of the answer is now provided as follows:

Given:

f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)

R_{3} = (x) = (e^z /4!)x^4

Since the aim is R_{3}(x) < 0.001, this implies that:

(e^z /4!)x^4 < 0.0001 ………………………………….. (2)

Multiply both sided of equation (2) by (1), we have:

e^4x^4 < 0.024 ……………………….......……………. (4)

Taking 4th root of both sided of equation (4), we have:

|xe^(z/4) < 0.3936 ……………………..........…………(5)

Dividing both sides of equation (5) by e^(z/4) gives us:

|x| < 0.3936 / e^(z/4) ……………….................…… (6)

In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:

|x| < 0.3936 -----> 0 < x < 0.3936

Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.

3 0
3 years ago
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