<p>I need help with review on these questions so could you help break it down and help me understand it.
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I also want to know how you read the unit circle like cos is the x values and sin is y. It's been a while.
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<p>The absolute min and max occur when the first derivative equals zero, but since this problem gives certain conditions of [-2, 3] we have to check those points as well, lets start with setting the first derivative to 0.</p>
<p>Lets use U substitution here and say (x^2 -2) = u
so y = u^(2/3), the differential would be dy = (2/3)u^(-1/3)du
du = 2xdx, since we want f’(x) aka dy/dx, lets start plugging back in the values of u in terms of x, so
dy = (2/3)<em>(x^2 - 2)^(-1/3)</em>2xdx, now divide by dx, so (dy/dx) = (4/3)<em>x</em>(x^2 - 2)^(-1/3)
or, f’(x) = (4/3)<em>x</em>(x^2 - 2)^(-1/3), now we set this equal to zero, and solve for x. I’m getting that x equals zero when 0 = (4/3)<em>x</em>(x^2 - 2)^(-1/3), now to determine if this is a relative max or min we could do the first derivative test, but that isn’t necessary because they gave us the conditions of [-2, 3], so to determine the absolute max and min we just plug in 0, -2, and 3 back into the original function, whichever one is the highest is the absolute max, whichever one is the min is the absolute min, and the middle one is irrelevant.</p>
<p>For absolute max and min coordinates I’m getting:
(-2, 4^(1/3)) min
(0, 4^(1/3)) interestingly enough its the same, it is also a min
(3, 49^(1/3)) max</p>
<p>I wouldn’t be surprised if I made a mistake in computation somewhere but the idea is valid, do you understand?</p>
<p>Remember that the slope is basically the same as the first derivative, so your trying to look for the smallest possible value of the derivative, you find this value by taking the second derivative setting it equal to zero and solve for x, then you plug this x back into the first derivative and by doing so you have found the smallest possible slope. So,</p>
<p>f(x) = x^3 - 3x^2 + 5x -1
f’(x) = 3x^2 - 6x + 5
f"(x) = 6x - 6
0 = 6x - 6, is when x = 1 so put this back into f’(x), so
f’(1) = 3(1)^2 - 6*1 +5
f’(1) = 3 - 6 + 5
f’(1) = 2
so 2 is smallest possible slope, but to check whether this isn’t the largest possible slope we plug in any other x value for f’(x) see if it is bigger or smaller that 2, and we know whether or not 2 is the smallest possible slope or greatest possible slope. It is the smallest possible slope.</p>
<p>It gives you the velocity function, so you do first derivative set it equal to zero, use the x values that you get from that and plug those x values back into the original function, you’ll also have to plug in 0 and 3, and then just look at which one is the largest functional value and which one is the smallest functional value.</p>
<p>It gave you the velocity function. Acceleration is the rate of change of velocity, so the first step is to do the first derivative. Once you do that you have the acceleration function, so know you have to do the derivative again, set it equal to zero, solve for x, plug in the values you got for x and 0 and 3 back into the acceleration function see which one is the smallest functional value and see which one is the largest functional value.</p>
<p>I get the last 2 thanks. But can you recheck your first steps work. I’m looking at it and I don’t think you did anything wrong.</p>
<p>What do you mean, for the first problem? Did I get it wrong or something?</p>
<p>You said you might have made a mistake but I don’t see a place for error.</p>
<p>I don’t see a mistake, but it would help knowing whether or not I got the right answer, do you have an answer key?</p>