we have a maximum at t = 0, where the maximum is y = 30.
We have a minimum at t = -1 and t = 1, where the minimum is y = 20.
<h3>
How to find the maximums and minimums?</h3>
These are given by the zeros of the first derivation.
In this case, the function is:
w(t) = 10t^4 - 20t^2 + 30.
The first derivation is:
w'(t) = 4*10t^3 - 2*20t
w'(t) = 40t^3 - 40t
The zeros are:
0 = 40t^3 - 40t
We can rewrite this as:
0 = t*(40t^2 - 40)
So one zero is at t = 0, the other two are given by:
0 = 40t^2 - 40
40/40 = t^2
±√1 = ±1 = t
So we have 3 roots:
t = -1, 0, 1
We can just evaluate the function in these 3 values to see which ones are maximums and minimums.
w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20
w(0) = 10*0^4 - 20*0^2 + 30 = 30
w(1) = 10*(1)^4 - 20*(1)^2 + 30 = 20
So we have a maximum at x = 0, where the maximum is y = 30.
We have a minimum at x = -1 and x = 1, where the minimum is y = 20.
If you want to learn more about maximization, you can read:
brainly.com/question/19819849
Answer:
2187
Step-by-step explanation:
× 3
729 × 3
2187
(x - 4)(x + 3)
= x^2 + 3x - 4x - 12
= x^2 - x - 12 (The answer is C)
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