PLEASE HELP ME SOLVE ### Determine Where the Given Function is Increasing and Where It is DecreasingGiven the function:\[ s(x) = -x^2 - 14x - 40 \]Select the correct intervals where the function is either increasing or decreasing from the following options:1. Decreasing on \((-\infty, -7)\), increasing on \((-7, \infty)\)2. Increasing on \((-\infty, \infty)\)3. Decreasing on \((-\infty, -7)\) and \((0, \infty)\), increasing on \((-7, 0)\)4. Increasing on \((-\infty, -7)\), decreasing on \((-7, \infty)\)### ExplanationTo determine the intervals of increase and decrease, find the critical points by first taking the derivative of the function \(s(x)\):\[ s'(x) = \frac{d}{dx}(-x^2 - 14x - 40) = -2x - 14 \]Set the derivative equal to zero to find the critical points:\[ -2x - 14 = 0 \]\[ -2x = 14 \]\[ x = -7 \]Now, test the intervals divided by the critical point \(x = -7\) to determine where the function is increasing and decreasing. The intervals to test are \((-\infty, -7)\) and \((-7, \infty)\):- For \(x \in (-\infty, -7)\): Choose a test point like \(x = -8\).\[ s'(-8) = -2(-8) - 14 = 16 - 14 = 2 \](Note: If the derivative is positive, the function is increasing.)- For \(x \in (-7, \infty)\): Choose a test point like \(x = 0\).\[ s'(0) = -2(0) - 14 = -14 \](Note: If the derivative is negative, the function is decreasing.)Thus, \(s(x)\) is decreasing on \((-\infty, -7)\) and increasing on \((-7, \infty)\).Based on this analysis, the correct choice is:***Decreasing on \((-\infty, -7)\), increasing on \((-7, \infty)\

Name: PLEASE HELP ME SOLVE ### Determine Where the Given Function is Increas
Uploaded: 2024-03-20T07:06:27.000Z
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Description: PLEASE HELP ME SOLVE ### Determine Where the Given Function is Increasing and Where It is DecreasingGiven the function:\[ s(x) = -x^2 - 14x - 40 \]Select the correct intervals where the function is either increasing or decreasing from the following o