prove that the curve 4y-x^2-8=0 is increasing in (0,4)?
回答 (2)
2016-07-25 6:13 am
✔ 最佳答案
4y - x² - 8 = 0(d/dx)(4y - x² - 8) = 0
4y' - 2x = 0
4y' = 2x
y' = x/2
y" = (d/dx)(x/2)
y" = 1/2
At (0, 4) :
y' = 0
y" = 1/2 > 0
Hence, the curve 4y - x² - 8 = 0 is increasing at (0, 4).
2016-07-25 6:32 am
4y − x² − 8 = 0
4y' − 2x = 0
4y' = 2x
y' = x/2
For x on interval (0, 4), y' = x/2 > 0, so y is increasing on this interval.
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