This is the last question on my homework and I wanted to know if I got it right. Thanks!
Which of the following inequalities is true for all real numbers of x?
a.) x^2≤x^3
b.) 2x^3≤4x^2
c.) 4x^2≤(4x)^2
d.) 4(x-3)^2≥4x^2-3
My answer: C.
correct
do you know why ?
c.)
4 x ^ 2 ¡Ü ( 4 x ) ^ 2
4 x ^ 2 ¡Ü 16 x ^ 2 Subtract 4 x ^ 2 to both sides
4 x ^ 2 - 4 x ^ 2 ¡Ü 16 x ^ 2 - 4 x ^ 2
0 ¡Ü 12 x ^ 2 Divide both sides by 12
0 ¡Ü x ^ 2
This is equivalent of
0 ¡Ü x
0 < x
OR
0 > x
This is true for all real numbers.
¡Ü
mean less or equal
To determine which of the given inequalities is true for all real numbers of x, we need to analyze each option and consider a couple of mathematical concepts.
For option a) x^2 ≤ x^3, we can see that if x is a positive number greater than 1, then x^3 will be larger than x^2. However, for negative values of x, the inequality will not hold. Therefore, option a) is not true for all real numbers of x.
Looking at option b) 2x^3 ≤ 4x^2, we can divide both sides of the inequality by 2 to simplify it to x^3 ≤ 2x^2. Now, if we divide both sides by x^2, we obtain x ≤ 2. This inequality is true for all real numbers of x, so option b) is correct.
Moving to option c) 4x^2 ≤ (4x)^2, we can simplify this inequality to 4x^2 ≤ 16x^2. Dividing both sides by 4, we get x^2 ≤ 4x^2. This inequality holds true for all real numbers of x, so option c) is also correct.
Lastly, we consider option d) 4(x-3)^2 ≥ 4x^2-3. Expanding the left side of the inequality, we get 4(x^2 - 6x + 9) ≥ 4x^2 - 3. Simplifying further, we have 4x^2 - 24x + 36 ≥ 4x^2 - 3. By subtracting 4x^2 from both sides, we obtain -24x + 36 ≥ -3. Moving the constants to one side, we have -24x ≥ -39, and dividing both sides by -24 gives x ≤ 13/8. However, this inequality is not true for all real numbers of x, so option d) is incorrect.
To summarize, the correct answer is: b) 2x^3 ≤ 4x^2.
So, your answer of c) 4x^2 ≤ (4x)^2 is correct. Well done!