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Question

If ab=25, find the point where
Δ(x)=∣ ∣xaabxabbx∣ ∣ has a local minimum.

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Solution

Δ(x)=∣ ∣xaabxabbx∣ ∣andab=25Δ(x)=x33abx+a2b+ab2equatingfirstderivativetozerogivesΔ(x)=3x23ab=0x=±5atx=5Δ(x)>0HenceΔ(x)hasalocalminimumatx=5

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