Solve for x. Round to the nearest tenth, if necessary. 3.4
1 answer:
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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' =
Frankly I like the first answer better, but you have a choice of both.
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' =
Frankly I like the first answer better, but you have a choice of both.
Answer:
The turning point of the graph is at (3,-16)
Step-by-step explanation:
The general equation of curve is given as:
y = x² + ax + b
The two point for which the equation satisfies is (0,-7) and (7,0).
Substitute (0,-7) in the general equation:
y = x² + ax + b
-7 = 0² + a(0) + b
b = -7
Substitute (7,0) in the general equation:
y = x² + ax + b
0 = 7² + 7a + b
Where b = -7
0 = 49 + 7a - 7
0 = 42 + 7a
a = -6
The equation of the curve is:
a = -6 and b = -7
y = x² + ax + b
y = x² - 6x - 7
Graph of the equation is attached below.
We can see that the the turning point of the graph is at (3,-16)
