Need help for 3 2/3 - 1/2
2 answers:
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For this case we have the following function:
The average rate of change is given by:
Evaluating the function for the given interval we have:
For x = 1.25:
For x = 2:
Then, replacing values we have:
Answer:
the average value of on the closed interval [1.25, 2] is:
The average rate of change is given by:
Evaluating the function for the given interval we have:
For x = 1.25:
For x = 2:
Then, replacing values we have:
Answer:
the average value of on the closed interval [1.25, 2] is:
Whenever you face the problem that deals with maxima or minima you should keep in mind that minima/maxima of a function is always a point where it's derivative is equal to zero.
To solve your problem we first need to find an equation of net benefits. Net benefits are expressed as a difference between total benefits and total cost. We can denote this function with B(y).
B(y)=b-c
B(y)=100y-18y²
Now that we have a net benefits function we need find it's derivate with respect to y.
Now we must find at which point this function is equal to zero.
0=100-36y
36y=100
y=2.8
Now that we know at which point our function reaches maxima we just plug that number back into our equation for net benefits and we get our answer.
B(2.8)=100(2.8)-18(2.8)²=138.88≈139.
One thing that always helps is to have your function graphed. It will give you a good insight into how your function behaves and allow you to identify minima/maxima points.
To solve your problem we first need to find an equation of net benefits. Net benefits are expressed as a difference between total benefits and total cost. We can denote this function with B(y).
B(y)=b-c
B(y)=100y-18y²
Now that we have a net benefits function we need find it's derivate with respect to y.
Now we must find at which point this function is equal to zero.
0=100-36y
36y=100
y=2.8
Now that we know at which point our function reaches maxima we just plug that number back into our equation for net benefits and we get our answer.
B(2.8)=100(2.8)-18(2.8)²=138.88≈139.
One thing that always helps is to have your function graphed. It will give you a good insight into how your function behaves and allow you to identify minima/maxima points.