Find the maxima and minima for y=x³-1.5x²
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Dy/ dx = 3x^2-3x
=3x(x-1)
Put dy/dx=0
So x=0 & x=1
Double derivative =6x-3
At x=oइस —3 thus maximum
At x=1is 3 thus minimum
=3x(x-1)
Put dy/dx=0
So x=0 & x=1
Double derivative =6x-3
At x=oइस —3 thus maximum
At x=1is 3 thus minimum
Answered by
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Explanation:
f(x) = x³ - 1.5x²
diff. both sides wrt x
f'(x)= 3x² - 3x
putting f'(x)= 0
3x² - 3x= 0
3x(x -1) = 0
x=1 , x= 0
Now again diff. , f"(x) = 6x - 3
Putting x=0, f"(x)= -3 is a point of local maxima
f(x) = 0³ - 1.5.0² = 0
Putting x=1, f"(x) = 3 is a point of local minima
f(x)= 1³ - 1.5.1² = -0.5
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