If y = 2x+8,then (X,y) is
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2x+8
then ,X=8-2
=6
I hope its work
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If y=2x−8y=2x−8
Then xy=x(2x−8)⇒xy=2x2−8xxy=x(2x−8)⇒xy=2x2−8x
Therefore, our new function is f(x)=2x2−8xf(x)=2x2−8x
Now, our new function has a minimum value at its first derivative zero:
f′(x)=ddx[2x2]−ddx[8x] f′(x)=2(2)x2−1−8(1)x1−1 f′(x)=4x−8 f′(x)=ddx[2x2]−ddx[8x] f′(x)=2(2)x2−1−8(1)x1−1 f′(x)=4x−8
So
0=4x−8⇔x=20=4x−8⇔x=2
Then, our new function has a minimum value at x=2x=2 And, it is:
f(2)=2(2)2−8(2)
f(2)=−8⇒y=−8

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