Math, asked by patelpooja9833, 1 day ago

find the minimum value of 4x^2 + 1÷x

Answers

Answered by Bhavy922

0

Answer:

4x^2+1/x=0

4x^2+1=0*x

4x^2=0-1

x^2=-1/4

X=√-1/4

X=-1/2 will be the answer

Step-by-step explanation:

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