To determine the behavior of the function [tex]\( h(x) = -5\sqrt{x} \)[/tex], we need to consider its derivative. The derivative will help us understand whether the function is increasing or decreasing on specific intervals.
1. Find the derivative of [tex]\( h(x) \)[/tex]:
[tex]\[ h(x) = -5 \sqrt{x} \][/tex]
We use the power rule and the chain rule for differentiation:
[tex]\[ h(x) = -5 x^{1/2} \][/tex]
To find the derivative [tex]\( h'(x) \)[/tex]:
[tex]\[ h'(x) = -5 \cdot \frac{1}{2} x^{-1/2} \][/tex]
[tex]\[ h'(x) = -\frac{5}{2} x^{-1/2} \][/tex]
[tex]\[ h'(x) = -\frac{5}{2\sqrt{x}} \][/tex]
2. Determine the sign of the derivative [tex]\( h'(x) \)[/tex] on the interval [tex]\( (0, \infty) \)[/tex]:
For [tex]\( x > 0 \)[/tex]:
[tex]\[ \sqrt{x} > 0 \][/tex]
Thus:
[tex]\[ h'(x) = -\frac{5}{2\sqrt{x}} \][/tex]
Since [tex]\( \sqrt{x} \)[/tex] is always positive for [tex]\( x > 0 \)[/tex], we see that [tex]\( -\frac{5}{2\sqrt{x}} \)[/tex] is always negative for [tex]\( x > 0 \)[/tex]. A negative derivative means that the function [tex]\( h(x) \)[/tex] is decreasing in the interval where the derivative is negative.
3. Consider the intervals given in the problem:
- The function [tex]\( h(x) = -5\sqrt{x} \)[/tex] is not defined for [tex]\( x < 0 \)[/tex] because the square root of a negative number is not a real number. Hence, the intervals [tex]\( (-\infty, 0) \)[/tex] are irrelevant.
- For [tex]\( x > 0 \)[/tex], the derivative [tex]\( h'(x) \)[/tex] is negative, which means the function is decreasing on [tex]\( (0, \infty) \)[/tex].
Therefore, the correct statement is:
[tex]\[ \boxed{\text{The function is decreasing on the interval } (0, \infty).} \][/tex]
