<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>I’m not giving you direct answers, but for the derivative cos(pix), use the chain rule, and note that the pi is a constant. Set the derivative equal to 0 to find your critical numbers.</p>
<p>And I’ve never heard of something raised to the plus exponent (ie. your example). There is such as thing as lim->3+ though.</p>
<p>Derivative: Found using Chain Rule</p>
<p>Note: π is pi</p>
<p>cos (πx) –> Outside is cosine, Inside is πx</p>
<p>So, take the derivative of the outside while leaving the inside alone: -sin (πx)
Then, multiply that by the derivative of the inside: π</p>
<p>So, the derivative of cos (πx) is -πsin (πx)</p>
<p>To find critical numbers, set the derivative equal to 0.</p>
<p>-πsin(πx) = 0
Divide away the -π, and you’re left with sin (πx) = 0
Temporarily replace πx with u, and solve the equation: sin u = 0
u would equal 0, π, 2π, etc.
But, we’re looking for x, and not u. These variables are related in that u = πx
So, u = πx = 0, π, 2π
To find x, divide out the π, and x (critical #s) is 0, 1, 2, etc.
The only critical number in your window [0, 1/6] is x=0.</p>
<p>However, I don’t know if you can have a critical number at an endpoint.</p>
<p>I’ve not heard of the plus in the exponent piece either.</p>
<p>Yes, you can have a critical point at an endpoint of a closed interval. Some texts define all endpoints of closed intervals to be critical points, while other texts only allow critical points to be locations where the first derivative function is equal to zero or is undefined.</p>
<p>This is not really a Calculus problem but while finding the y intercept for a problem that asks to analyze the graph, I stumbled on a small problem. The numerator is x^3 + x + 4. How do I find the value for x that would set the equation = 0? T_T</p>
<p>What’s the whole function?</p>
<p>(x^3 + x + 4) / (x^2 + 1) I think the denominator is irrelevant. The answer is -1.3788, I don’t know how to get that algebraically.</p>
<p>man, solving cubic equations is a pain in the butt.</p>
<p>Yeah, I can’t figure out how to solve that algebraically.</p>
<p>i can do it on paper but it’s kind of hard to explain esp. on a message board, try googling it</p>
<p>Is Newton’s method the only way to do it? =(</p>
<p>no, there’s thing called Cardano’s method which solves it using the “depressed cubic”</p>
<p>or you can always have WolframAlpha solve it for you and then check the steps ;)</p>
<p>There’s also a cubic formula that’s very much like the quadratic formula, but I don’t recommend anybody actually learn the cubic formula, as it’s ever so slightly obnoxious. ;-)</p>
<p>TheMathProf is right. The cubic formula is obnoxious. I just looked it up on wikipedia. O.O</p>
<p>Do people actually use that monstrosity?</p>
<p>Nooooo they do not. It’s an interesting result only because it exists (there are no explicit formulas for the roots of any higher order polynomials).</p>