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(4pts) Show how either the first derivative test or the second derivative test shows that - minimum: this point is a you use first derivative test, make sign chart)...

Question

(4pts) Show how either the first derivative test or the second derivative test shows that - minimum: this point is a you use first derivative test, make sign chart)

(4pts) Show how either the first derivative test or the second derivative test shows that - minimum: this point is a you use first derivative test, make sign chart)



Answers

Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum. $$h(y)=y^{2}-\frac{1}{y}$$

From 1935. So given why prime, is it gonna one minus either X? Why? Double problems equal a negative view of the X is that why prime you go to zero and therefore one to go to you? The ax for access equals zero. Then we find out what the value is at X equals zero. That's a good negative wine, which is less than zero. And that way we know that it's gonna be concave down. So it's a maximum point. Therefore, the maximum points at zero comma, negative one.

Okay now we've got dysfunction and we have to take the derivative Using the quotient rule. It's the denominator times the derivative of the numerator which is just three minus numerator. Three X plus one times the derivative of the denominator which is two X. Over denominator squared. Simplifying three X squared minus six X squared would be negative three X squared. And then I've got a minus two X. Two X. And I've got a positive three. Okay. Okay um Numerator equal zero. So that's gonna be negative three X squared minus two X plus three equals zero. Now doing the quadratic formula negative B plus or minus the square root of B squared minus four. A. See over to Okay two plus or minus the square root. That's four plus 36. Which is 40 over negative six. Taking a two out of everything that's going to be one plus or minus. You know I'm going to put the negative in the numerator negative one plus or minus the square root of 10. They cancelled out a square to four over three. So that means that it's either going to be negative one minus the square root of 10 over three or negative one plus the square root of 10 over three. So now I need to take the second derivative. So again with the product role denominator which is X squared plus one squared times the derivative of the numerator which would be did not find the denominator derivative of the numerator is going to give me negative six X minus two minus the numerator times the derivative of the denominator which is going to be too times X squared plus one driven of the inside is two X over denominator squared X squared plus one to the fourth power. Now an X squared plus one is going to cancel out. Okay, now we simplify X squared minus times negative six X squared is going to be negative six X cubed. Uh huh. Okay. Uh minus two X squared minus six X six, yep minus two. Just double checking my work here. Okay now I need to take a negative four X times everything. The denominator is always going to be positive because it's squared. I'm just noticing that right now so I really just have to worry about whether the numerator is positive or negative. So simplifying some more. Okay, so now I need to substitute in those numbers into negative six X cubed plus 10 X squared plus two X minus 14. All right, I'm gonna go to a calculator for that negative six and I'm just going to write X cubed plus 10 X squared plus two X minus 14. Yeah. And then for X I'm gonna put in negative one minus the square to 10/3. Negative one minus square root of 10 over 3/3. Okay, So I'm going to substitute that in. I'm actually using DeZ mose. Unfortunately, I didn't set this up for you to see it copy copy copy. Okay, so uh for the first one I get positive 18, so that means positive curvature is like this. So that would be a minimum for the second one. Plus plus whips. I get negative 9.6. So that's going to be like this, so that's going to be a maximum and that's correct. Thank.

Now we have this function Z. There, I'm going to use the quotient role, denominator times derivative of the numerator minus numerator times derivative of the denominator, which is to Z over denominator squared. Simplify that. Let's see. Two Z cubed minus two Z cube. Those will cancel out. And in the numerator I'm just left with two Z. Mhm Over one plus Z squared. And if the numerator equal zero Within z equals zero second derivative, denominator times derivative of the numerator, which is just to minus numerator times derivative of the denominator, which is to Times one plus Z over denominator squared ah Squared again. So that's going to be the 4th power. I see that there's a one plus Z in every term and in the denominator. So I'm going to cancel that out. And that gives me just simplifying. Actually I'm going to distribute uh to z minus four Z. That's going to give me -2 Z. And then I just have a two left, one plus Z, 3rd power. Nothing else cancels out. So if negative whoops now I need to find G double prime of zero. That's going to give me too over one, which is two, which is positive, so that's going to be a minima positive means upward curvature. Thank you for watching.

For this problem we are asked to find the critical points in the intervals on which the function Y equals X squared times eat our X is increasing or decreasing. Were then asked to use the first derivative test to determine whether the critical point yields a local men or max or neither. So to begin, we want to take the derivative with respect to X. So we'll need to apply the product rule giving us two X. E. To the power of X plus X squared E. To the power of X. We want to set that equal to zero. So what we can do is factor out the E to the power of X. Actually we can factor out an X. E. To the power of X. And then we get X plus two or exit with our backs times X plus two. And that needs to equal zero. You can see that that's going to be zero when X equals zero or when X equals negative two. Now we can partition the real numbers so we have zero. Actually we should have negative too and then zero to the right of zero X. E. To the power of X will be positive exposed to will be positive. So the derivative will be positive Between -2 and zero executing power of X will be negative and X plus two will be positive. So we'll have negative times positive giving us negative And then to the left of -2, excessive power backs will be negative and X plus two will be negative, we'll have double negative so we'll be positive. Now that means that at negative two we're going from increasing decreasing. So that must be a max and at X equals zero. We're going from decreasing to increasing. So that must be a min having that. Then you can say that our interval of increase is going to be from negative infinity up to negative two, Union with zero up to infinity, and our interval of decrease is going to be from negative to up to zero.


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