y=x^2 +2x+5 ,prove that minma exist for this curve and find that minima
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dy/dx = 2x + 2
For max or min
dy/dx = 0
2x + 2 = 0
x = - 1
d^2y/dx^2 = 2 > 0
Therefore y is min at x = -1
And minimum value is 1 - 2 + 5 = 4
For max or min
dy/dx = 0
2x + 2 = 0
x = - 1
d^2y/dx^2 = 2 > 0
Therefore y is min at x = -1
And minimum value is 1 - 2 + 5 = 4
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