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Question : 62 of 160
Marks:
+1,
-0
Solution:
Consider the given equation.
y=x3−3x2+5‌quad... Differentiate the equation with respect to
x ‌‌=3x2−6x For local maxima and minima,
‌‌=0 3x2−6x=0 3x(x−2)=0 x=0,2 Differentiate equation (2) with respect to
x ‌‌=6x−6 (‌‌)x=0=−6<0 Therefore,
x=0 is a point of local maxima.
(‌‌)x=2 =6×2−6 =6>0 Therefore,
x=2 is a point of local minima.
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