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The length of the function y = 3x over the given interval [0, 2] is 3.2 units
For given question,
We have been given a function y = 3x
We need to find the length of the function on the interval x = 0 to x = 2.
Let f(x) = 3x where f(x) = y
We have f'(x) = 3, so [f'(x)]² = 9.
Then the arc length is given by,
This means, the arc length is 3.2 units.
Therefore, the length of the function y = 3x over the given interval [0, 2] is 3.2 units
Learn more about the arc length here:
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Answer:
The solution in interval notation is:
.
The solution in inequality notation is:
.
Step-by-step explanation:
I think you are asking how to solve this for .
Keep in mind .
If then
.
Subtract on both sides:
Factor the difference of squares :
Simplify inside the factors:
The left hand side is a parabola that faces up. I know this because the degree is 2.
The zeros of the the parabola are at x=-6 and x=2/5.
We can solve x+6=0 and 5x-2=0 to reach that conclusion.
x+6=0
Subtract 6 on both sides:
x=-6
5x-2=0
Add 2 on both sides:
5x=2
Divide both sides by 5:
x=2/5
Since the parabola faces us and then we are looking at the interval from x=-6 to x=2/5 as our solution. That part is where the parabola is below the x-axis. We are looking for where it is below since it says the where is the parabola<0.
The solution in interval notation is:
.
The solution in inequality notation is:
.