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Find and classify the relative extrema of f(") 21V' %r +))+1...

Question

Find and classify the relative extrema of f(") 21V' %r +))+1

Find and classify the relative extrema of f(") 21V' %r +))+1



Answers

Find the relative extrema of $f | S$. $$\begin{array}{l}f: \mathbb{R}^{3} \rightarrow \mathbb{R},(x, y, z) \mapsto x^{2}+y^{2}+z^{2}, S=\{(x, y, z) | \left.z \geq 2+x^{2}+y^{2}\right\}\end{array}$$

Okay. So for this example were given a function F. That maps from our to to our consisted of X. M. Lies and can be expressed as X squared plus Y squared. Were also given a sad S that consists of excess and wise such that Y is greater than or equal to two. So we need to find the relative extreme of F. That's restricted to s. All right. So let's get started for sinks first. Let's just finally gradient of F. And that is just two X. And two. Why? And let's say A equals zero from this. We see that X not is equal zero and why not is equal to zero? However, this is not allowed. Okay. So therefore um the minimum value ah will occur A at X not is equal to zero and why? No, it is equal to two. Okay. So thus the minimum value of our function that's of X. Y. Is so F of X comma Y is equal to zero Plus two squared which is equal to four, sir? Answer

So here we want to use any method to find the relative extreme A. So we are going to start by looking at their backs equals our graph is a function of X times x minus one squared. Oh yeah. So with this in mind let's graph RFP innovex function. We know that we could solve this just by foiling south. That would be X squared minus two X plus one multiply everything by X through. I think that's the easiest way because then we get into polynomial form and then we can just differentiate easily using the power rule. So with that in mind we graph the first derivative graph when we get to critical points one third and one. So what this tells us is that since the graph is increasing in slope and then decreasing slope, Then this critical point right here at 1/3 is going to be a relative maximum. And then this point right here one we're decreasing in slope and then we reached a critical point and then we start increasing in slope. So that tells us that the .1 is going to be a relative minimum.

With this graph, we see that there is a relative maximum at negative 1.135 and then there's a relative minimum at 00

Presuming on this craft, furious that there's a relative minimum on. And if we were to do this by taking the derivative and all that, and, um, keeping it in terms of the actual answer, this point right here is equivalent to negative natural log of two and a negative 1/4.


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