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Problem 53

a. Find the local extrema of each function on the…

05:45

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Problem 52

a. Identify the function's local extreme values in the given domain, and say where they occur.
b. Which of the extreme values, if any, are absolute?
c. Support your findings with a graphing calculator or computer grapher.
$$
g(x)=\frac{x^{2}}{4-x^{2}}, \quad-2< x \leq 1
$$

Answer

$$
\begin{array}{l}{\text { (a) local minimum at } x=2-\sqrt{3}(b) \text { (b) absolute minimum at } x=2} \\ {\text { c) See the graph }} \\ \hline\end{array}
$$


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Video Transcript

All right. So huge CI of ax is X squared over four minus x squared on the domain negative to less and act celestron equal to one. If we take the derivative, we'LL get it or minus X squared times two X and then it's going to end up being plus minus position to be a negative X squared times negative two axes becomes positive over four minus X squared squared. Good. So when is uh let's see when is cheap prime going to be zero We'LL win this numerator zero see? So this is negative two x cubed plus to execute So that's actually going to go to Syria. We're just going to be left with eight x under. So we have eight x equals zero or when x zero And then when is ji undefined? She promised to find well, when four minus X squared zero or ex is plus or minus two. But now negative too is not in the domain. So we're only going to consider politics. Okay, so let's look at oh, can actually Ah, Positive too is not in the domain either, so we don't consider either one of us. Okay, So X equals zero is on the critical point, though we do have an end point at one. So here's negative to or not. I'm putting in a bid to Here's one and then you have zero to the left of zero. The denominator here is always positive, so that's gonna be negative. Positive. So we have a local men and zero in a value of zero. And this is actually actually the absolute men zero, because it's decreasing before and increasing after. And then we have a local max at X equals one with the value of looks like one third. Good. And now we might say, OK, this is the absolute max, but we need to see what's happening as extra purchase. Negative two from the right. So noticed that the limit his expressions. Thank you too. Uh, right. If she of X is going to, well, insanity so there's gonna be no absolute max because dysfunctions going off to infinity is extra purchase Negative two from the right

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