identify the open intervals on which the function is increasing or decreasing. (enter your answers using…

Explanation:

Step1: Find the derivative of the function

The derivative of (h(x) = 75x - x^{3}) is (h'(x)=75 - 3x^{2}) (using the power rule ((x^{n})'=nx^{n - 1})).

Step2: Set the derivative equal to zero and solve for (x)

Set (h'(x)=0), so (75 - 3x^{2}=0). First, factor out a (3): (3(25 - x^{2})=0), then (25 - x^{2}=0). Using the difference - of - squares formula (a^{2}-b^{2}=(a + b)(a - b)) with (a = 5) and (b=x), we have ((5 + x)(5 - x)=0). Solving ((5 + x)(5 - x)=0) gives (x=-5) or (x = 5).

Step3: Test intervals

We have three intervals to test: ((-\infty,-5)), ((-5,5)), and ((5,\infty)).

• For the interval ((-\infty,-5)), let (x=-6). Then (h'(-6)=75-3\times(-6)^{2}=75 - 108=-33<0).
• For the interval ((-5,5)), let (x = 0). Then (h'(0)=75-3\times0^{2}=75>0).
• For the interval ((5,\infty)), let (x = 6). Then (h'(6)=75-3\times6^{2}=75 - 108=-33<0).

Answer:

increasing: ((-5,5)) decreasing: ((-\infty,-5)\cup(5,\infty))