Determine the values of xfor which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001.(Enter your answer using interval notation. Round your answer to four decimal places.)

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)

= (x) = (e^z /4!)x^4

Since the aim is (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.

option: d is the correct answer.

d) none of the above.

step-by-step explanation:

clearly from the figure attached to the question we could see that there is no shape given. hence without any shape given in the figure and after rotation we can't obtained a square pyramid.

since, to obtain a image we need a proper figure of the pre-image so that it can be transformed.

hence, the correct answer is:

d) none of the above.

step-by-step explanation: