Question

Consider the differential equation \frac{dx}{dt} = (x - 1)^2(x - 5)(6 - x) The smallest critical value is x = The solution curve for the smallest critical value is Select an answer The middle critical value is x = Its solution curve is Select an answer The largest critical value is x = Its solution curve is Select an answer

          Consider the differential equation \frac{dx}{dt} = (x - 1)^2(x - 5)(6 - x)
The smallest critical value is x = 
The solution curve for the smallest critical value is Select an answer
The middle critical value is x = 
Its solution curve is Select an answer
The largest critical value is x = 
Its solution curve is Select an answer
        
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Consider the differential equation (dx)/(dt) = (x - 1)^2(x - 5)(6 - x)
The smallest critical value is x = 
The solution curve for the smallest critical value is Select an answer
The middle critical value is x = 
Its solution curve is Select an answer
The largest critical value is x = 
Its solution curve is Select an answer

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Consider the differential equation dx = x - 12(x - 5(6 - x)) dt. The smallest critical value is x = . The solution curve for the smallest critical value is . The middle critical value is x = . Its solution curve is . The largest critical value is x = . Its solution curve is .
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Transcript

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00:01 Okay, so we need to do the second derivative test for this x e to the negative x squared.
00:08 And what we need to do is first of all find the derivative, which is a product rule.
00:13 There's a product right here.
00:14 So i like to do the derivative of x, leave e to the negative x squared alone.
00:18 And now you leave the x alone, you do the derivative of e to the negative x, which is itself.
00:24 But then you have to multiply by the derivative of negative x squared, which would be negative 2x.
00:29 So what i would do is factor out this e to the negative x squared.
00:34 You're left with 1 minus 2x squared in there.
00:39 Now this little bit will always be positive, so we don't have to worry about that.
00:44 But 1 minus 2x squared could equal zero.
00:48 So i can add that over, divide, and then square root.
00:53 Now i'll just leave it as positive or negative the square root of one half or 1 over root 2.
01:00 And then we need the second derivative.
01:04 So i'm going to look at my blue equation and do the second derivative, which is another product rule where i would leave the e to the negative x squared or take the derivative of that, negative 2x.
01:18 Chain rule again, leaving the right side alone.
01:23 And now leave e to the negative x squared alone, taking the derivative of the right side, which would be negative 4x.
01:29 And again, i can factor, but at this point i probably need to distribute.
01:36 Like negative 2x, if i add this negative 4x to it, i have negative 6x.
01:40 And then negative 2x times negative will be a positive 4x cubed.
01:47 And now what you have to do for the second derivative test is see what happens if you plug in the 1 over root 2.
01:55 Now this piece will be positive still.
01:59 And let's just go to a calculator.
02:01 Although, yeah, negative 6 times 1 over root 2 plus 4 times 1 over root 2 cubed gave me an answer of negative 2 .828...
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