Help me with please i will give a brainliest for a good answer
2 answers:
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Problem 1
With limits, you are looking to see what happens when x gets closer to some value. For example, as x gets closer to x = 2 (from the left and right side), then y is getting closer and closer to y = 1/2. Therefore the limiting value is 1/2
Another example: as x gets closer to x = 4 from the right hand side, the y value gets closer to y = 4. This y value is different if you approach x = 0 from the left side (y would approach y = 1/2)
Use examples like this and you'll get the results you see in "figure 1"
For any function values, you'll look for actual points on the graph. A point does not exist if there is an open circle. There is an open circle at x = 2 for instance, so that's why f(2) = UND. On the other hand, f(0) is defined and it is equal to 4 as the point (0,4) is on the function curve.
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Problem 2
This is basically an extension of problem 1. The same idea applies. See "figure 2" (in the attached images) for the answers.
With limits, you are looking to see what happens when x gets closer to some value. For example, as x gets closer to x = 2 (from the left and right side), then y is getting closer and closer to y = 1/2. Therefore the limiting value is 1/2
Another example: as x gets closer to x = 4 from the right hand side, the y value gets closer to y = 4. This y value is different if you approach x = 0 from the left side (y would approach y = 1/2)
Use examples like this and you'll get the results you see in "figure 1"
For any function values, you'll look for actual points on the graph. A point does not exist if there is an open circle. There is an open circle at x = 2 for instance, so that's why f(2) = UND. On the other hand, f(0) is defined and it is equal to 4 as the point (0,4) is on the function curve.
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Problem 2
This is basically an extension of problem 1. The same idea applies. See "figure 2" (in the attached images) for the answers.
Answer:
a) The tower is 90 feet tall
b) She reaches the bottom at t = 18 minutes.
c) Her speed at time t is 5 \sqrt[]{5} ft/minute
d) Her acceleration at time t is 10 ft/minute^2
Step-by-step explanation:
Consider the path described by the child as going down the tower to have the following parametrization
a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.
b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.
c) Given the parametrization we are given, the velocity of the child at time t is given by . The speed is defined as the norm of the velocity vector,
so, the speed at time t is given by
d) ON the same fashion we want to know the norm of the second derivative of .
We have that so the acceleration is given by