Please help with this problem. Thanks

a/ Show that for any real number x,
if x>1 then |x^4|=< |23x^4 + 8x^2 + 4x|

b/ Show that for any real number x,
if x>1 then |23x^4 + 8x^2 + 4x| => |x^4|

c/ Use the big Theta and big O-notations to express the results of part (a) and (b)

d/ What can you deduce about the order of 23x^4 + 8x^2 + 4x ?

a/ To show that for any real number x, if x > 1 then |x^4| <= |23x^4 + 8x^2 + 4x|, we can start by simplifying the expression |23x^4 + 8x^2 + 4x|.

First, let's factor out x from each term:
|23x^4 + 8x^2 + 4x| = |x(23x^3 + 8x + 4)|

Since x > 1, we can conclude that x^4 > x and x^4 > 1. Therefore, we have:
|x^4| > |x| > 1

Now, let's consider the expression inside the absolute value |23x^3 + 8x + 4|. We know that x > 1, so each term in the expression is positive. Therefore, we can drop the absolute value signs:
23x^3 + 8x + 4

Next, we can compare the magnitudes of the terms x^4 and 23x^3 + 8x + 4:
|x^4| <= 23x^4 since 23x^4 is always greater than x^4 when x > 1

Now, let's substitute this inequality into the original expression:
|x^4| <= |23x^4 + 8x^2 + 4x|
|x^4| <= 23x^4 + 8x^2 + 4x

Thus, we have shown that for any real number x, if x > 1, then |x^4| <= |23x^4 + 8x^2 + 4x|.

b/ To show that for any real number x, if x > 1 then |23x^4 + 8x^2 + 4x| >= |x^4|, we can use a similar approach.

We can start by simplifying the expression |23x^4 + 8x^2 + 4x| again.
|23x^4 + 8x^2 + 4x| = |x(23x^3 + 8x + 4)|

Since x > 1, we know that x^4 > x and x^4 > 1. Therefore, we have:
|x^4| > |x| > 1

Now, let's consider the expression inside the absolute value |23x^3 + 8x + 4|. Again, we can drop the absolute value signs since each term is positive:
23x^3 + 8x + 4

Next, we can compare the magnitudes of the terms x^4 and 23x^3 + 8x + 4:
|x^4| <= 23x^4 since 23x^4 is always greater than x^4 when x > 1

Now, let's substitute this inequality into the original expression:
|x^4| <= |23x^4 + 8x^2 + 4x|
|x^4| <= 23x^4 + 8x^2 + 4x

Thus, we have shown that for any real number x, if x > 1, then |23x^4 + 8x^2 + 4x| >= |x^4|.

c/ Now let's express the results of part (a) and (b) using the big Theta and big O-notations.

In part (a), we showed that |x^4| <= |23x^4 + 8x^2 + 4x|. This implies that |23x^4 + 8x^2 + 4x| is an upper bound for |x^4|. Using big O-notation, we can express this as:
|x^4| = O(|23x^4 + 8x^2 + 4x|)

In part (b), we showed that |x^4| <= |23x^4 + 8x^2 + 4x|. This implies that |x^4| is a lower bound for |23x^4 + 8x^2 + 4x|. Using big Theta-notation, we can express this as:
|x^4| = Θ(|23x^4 + 8x^2 + 4x|)

d/ From the results of part (a) and (b), we can deduce that the order of the polynomial 23x^4 + 8x^2 + 4x is O(x^4) and Θ(x^4). This means that as x approaches infinity, the dominant term is x^4.