f(x)=3(x−21)2+4 (1)
Differentiating above equation with respect to x we get
f′(x)=6(x−21)×1+0 (2)
Put f′(x)=0 in above equation
⇒6(x−21)=0
⇒x=21
Put x=21 in equation (1), we get
⇒f(21)=4
Differentiating eq.(2) with respect to x,
f′′(x)=6>0 (local minima condition)
∴ x=21 gives the minimum value of the function f(x)
f(x) = 4 is the minimum value of the function.
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