To determine which of the given functions is always increasing over its entire domain, we need to examine their derivatives. A function is increasing where its derivative is positive.
Let’s analyze each function:
1. Function \( f(x) = x^2 \)
- Derivative: \( f'(x) = 2x \)
- For \( f'(x) > 0 \), we need \( 2x > 0 \)
- This is true when \( x > 0 \)
- Hence, \( f(x) = x^2 \) is not increasing over its entire domain (it only increases for \( x > 0 \), and decreases for \( x < 0 \)).
2. Function \( f(x) = x^3 - 4x \)
- Derivative: \( f'(x) = 3x^2 - 4 \)
- For \( f'(x) > 0 \), we need \( 3x^2 - 4 > 0 \)
- \( 3x^2 > 4 \)
- \( x^2 > \frac{4}{3} \)
- \( x > \sqrt{\frac{4}{3}} \) or \( x < -\sqrt{\frac{4}{3}} \)
- Hence, \( f(x) = x^3 - 4x \) is not increasing over its entire domain (it increases when \( |x| > \sqrt{\frac{4}{3}} \) and decreases when \( |x| < \sqrt{\frac{4}{3}} \)).
3. Function \( f(x) = -2x + 5 \)
- Derivative: \( f'(x) = -2 \)
- \( f'(x) = -2 \) is always less than 0
- Hence, \( f(x) = -2x + 5 \) is always decreasing.
4. Function \( f(x) = \sqrt{x + 2} \)
- Derivative: \( f'(x) = \frac{1}{2\sqrt{x + 2}} \)
- The derivative, \( \frac{1}{2\sqrt{x + 2}} \), is always positive for all \( x \geq -2 \)
- This is because the square root of a number is always positive, ensuring the expression under the division is positive.
- Hence, \( f(x) = \sqrt{x+2} \) is increasing over its entire domain \( [ -2, ∞ ) \).
Conclusion: Among the given functions, the function [tex]\( f(x) = \sqrt{x+2} \)[/tex] is always increasing over its entire domain.
