Small calculus question

<p>What would the derivative be for and the critical numbers be for cos(pix)with endpoints [0,1/6]? And what does it mean when something is raised to the plus exponent like x^3+?</p>

<p>y = -pi (sin(pi*x))</p>

<p>In the range [0, 1/6], it has a y-intercept at x=0. It should be the only critical point as the point where the derivative is zero is at x=1/2, which is out of your range. </p>

<p>It’s fairly standard. The derivative of cos(kx) is -k sin(kx). </p>

<p>I’m not sure what you mean by your second question. What do you mean by “something” being raised to a positive exponent? You’ll have to clarify what you mean.</p>

<p>Wait so the critical numbers for that derivative is 1/2? I need help finding that. When you set that derivative equal to zero. Also it’s like x^to a power like 3 with a plus sign after it.</p>

<p>Well, I gave you the critical number(s) for the derivative, the sine function. Obviously, it is different for the original cosine function. </p>

<p>What are you asking about x^3?</p>

<p>I assume its referring to all monomials with exponents 3 or greater.</p>

<p>For example, x^3+ may include x^3, x^4, x^5, x^6,…</p>

<p>Another way to represent that is x^n where n >= 3</p>