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Setler [38]
3 years ago
7

Calculus question?

Mathematics
1 answer:
Ann [662]3 years ago
5 0
Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)

Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2   Now differentiate that. It should be much easier.

Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x  I wonder if there's anything else you can do to this. If there is, I don't see it.

I suppose this is possible.
y' = 3/x^(1/2) + 6x

y' = \frac{3 + 6x^{3/2}}{x^{1/2}}

Frankly I like the first answer better, but you have a choice of both.
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The solution set of the inequality 2x − y &gt; 2 consists of all the points __ ? the line __ ?
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7 0
3 years ago
find the x-intercepts of the parabola with vertex(6,27) and y-intercept (0,-81). write answer in form (x1,y1)(x2,y2). if necessa
sergey [27]
The vertex form of the equation is:
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- 81 = a ( 0 - 6 )² + 27
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a = - 108 : 36
a = - 3
y = - 3 ( x - 6 )² + 27
y = - 3 ( x² - 12 x + 36 ) + 27
y = - 3 x² + 36 x - 108 + 27
y = - 3 x² + 36 x - 81
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x 1 = 3,  x 2 = 9
Answer:
The x-intercepts are ( 3, 0 ) and ( 9, 0 ).
7 0
3 years ago
Read 2 more answers
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