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critical point of 4x+4sin(x)
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Derivative set equal to zero.
4 +4cos(x) =0 cos(x) = -1 x= (2n-1)Pi n= 1,2,......n
i got zero when i did that but its online homework and its not accepting it
differentiating = 4 + 4 cos x at a critical point 4 + 4 cos x = 0 cos x = -1 x is what myko said
try puting\[(2n-1)\pi\]
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in degrees x = -180, - 540 and so on
or plus x = 180, 540
inflection point would be the second derivative equal to zero
Answer would be Pi/2
Cos(x)=-1 X would be 180 degrees
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for the inflection point?
Second derivative set equal to zero and solve
is the second derivative -4sin(x)
?
Yes, that's correct.
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is the critical point 0?
Yes that's it
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