What is the image point of (2, 4) after a translation left 4 units and down 1 unit?
1 answer:
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Answer:
(A)The absolute maximum of f(x) is at x=5
Step-by-step explanation:
F(x)=−2x³+21x²−60x on [1,6]
Find the first derivative.
f'(x)=-6x²+42x-60
Set the first derivative equal to zero.
-6x²+42x-60=0
Solve to find the critical points.
x=2,5
Next, we use the endpoints and all critical points on the interval to test for any absolute extrema over the given interval.
x=1,2,5,6
This is done by evaluating the function at x=1,2,5,6.
F(1)=−2(1)³+21(1)²−60(1)=-41
F(2)=−2(2)³+21(2)²−60(2)=-52
F(5)=−2(5)³+21(5)²−60(5)=-25
F(6)=−2(6)³+21(6)²−60(6)=-36
The maximum will occur at the highest f(x) value and the minimum will occur at the lowest f(x) value.
Absolute Maximum: (5,-25)
Absolute Minimum: (2,-52)
The graph is attached.
The absolute minimum and maximum of f(x) are (2,-52) and (5,-25) respectively.
