Let f(x)=x2+10x+29. What is the minimum value of the function?

asked Jun 10, 2024 203k views

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Answer:

Explanation:

$f(x)=x^2+10x+29 \\ \\ f(x)=(x^2+10x+25)+29+25 \\ \\ f(x)=(x+5)^2+54 \\ \\ \therefore (x+5)^2 \geq 0 \implies (x+5)^2+54=f(x) \geq 54$

Dinesh J answered Jun 15, 2024

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