home work help

Problem 2:

 

The expression we're given involves squares of several terms. Here's how to find the minimum value for all real numbers x:

 

Completing the Square: Notice that each term in the expression is a squared term of the form (x + a)^2. To minimize the expression, we can try to manipulate it into the form of a perfect square trinomial (a squared term plus a constant term).

 

Constant Term Issues: Unfortunately, we cannot directly complete the square for each term because there are constant terms (like 12, 7, etc.) within the parentheses. Completing the square typically involves adding and subtracting a constant term that depends on the coefficient of our x term. However, in this case, adding such a constant term inside the parentheses would also affect the squared term itself, making it no longer a perfect square.

 

Grouping and Common Factors: Instead of completing the square for each term individually, let's look for ways to group the terms and find common factors. We can rewrite the expression as:

 

(x + 12)^2 + (x + 7)^2 + (x + 3)^2 + (x - 4)^2 + (x - 8)^2 =

(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64)

 

Notice that each term except the first has a common factor of (x^2 + px + q), where p and q are constants depending on the specific term.

 

Factoring and Simplifying: Now we can factor out the common factors and group the remaining terms:

 

(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64) =

 

(x^2 + 12x + 144) + (x^2 + 14x + 49) + (x^2 + 6x + 9) + (x^2 - 8x + 16) + (x^2 - 16x + 64)

 

= (x^2) + (x^2) + (x^2) + (x^2) + (x^2) + (12x + 14x + 6x - 8x - 16x) + (144 + 49 + 9 + 16 + 64)

 

= 5(x^2) - 14x + 282

 

Non-Negative Term: The first term, 5(x^2), is always non-negative because any real number squared is non-negative. The minimum value it can take is 0, which occurs when x = 0.

 

Minimizing the Expression: The remaining term, -14x + 282, is minimized when -14x is maximized (since it's being subtracted). However, -14x is maximized when x is minimized (remember x is a real number).

 

Since we have no restrictions on the lower bound of x (it can be any real number), -14x can be infinitely negative as x approaches negative infinity.

 

Minimum Value: Therefore, the minimum value of the entire expression approaches negative infinity as x approaches negative infinity.

 

However, the question asks for the minimum value for ALL real numbers x. Since 5(x^2) is always non-negative and -14x can be arbitrarily negative, the minimum value of the expression is achieved when 5(x^2) = 0 (which occurs at x = 0) and -14x reaches its maximum negative value.

 

Answer: The minimum value of the expression is 282.

Problem 3:

 

We can solve this problem by strategically manipulating the inequalities and using our understanding of exponents. Here's how to approach it:

Break down the exponentials:

 

Notice that 2 raised to the power of 200 appears on the left side of the first inequality. We can rewrite it as (2^2)^100 which is equal to 4^100.

Simplify the inequalities:

 

With the simplification from step 1, the inequalities become: 4^100 < n^100 < (130n)^{50}

 

Taking the 50th root:

 

To compare the terms more effectively, let's take the 50th root of all three parts of the inequalities. This is a valid operation because all terms are positive. Remember that taking the nth root preserves the order of the inequality when dealing with positive numbers.

 

Applying the 50th root: 2 < n < √(130n)

 

Squaring both sides:

To get rid of the square root, we can square both sides of the inequality. However, squaring introduces a potential issue: squaring flips the direction of the inequality when the term being squared is negative.

 

Since n is a positive integer by definition, squaring both sides will preserve the direction of the inequality. So we have: 4 < n^2 < 130n

 

Analyze the terms:

 

The leftmost inequality (4 < n^2) tells us that n^2 must be greater than 4. This means n itself must be greater than 2 (since the square of a number cannot be less than the number itself).

 

The rightmost inequality (n^2 < 130n) tells us that n^2 is less than some multiple of n (specifically, 130 times n). This can only be true if n itself is less than 130.

 

Identify valid n:

 

So, to satisfy both inequalities, n must be greater than 2 and less than 130.

 

Now we consider all the positive integers within this range: 3, 4, 5, ..., 128, 129.

 

Counting valid integers:

 

There are a total of 129 - 3 + 1 = 127 positive integers that satisfy the conditions.

 

Answer: There are 127 positive integers n that satisfy the inequalities: 2^{200} < n^{100} < (130n)^{50}.