Question

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

$h(x)=75x - x^{3}$

increasing

decreasing

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