The minimum value of f(x) = x^2 + 4x + 5 is | vedclass

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(B) To find the minimum value of the quadratic function $f(x) = x^2 + 4x + 5$,we can complete the square or use the derivative method.Method $1$: Comp

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$1$

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(B) To find the minimum value of the quadratic function $f(x) = x^2 + 4x + 5$,we can complete the square or use the derivative method.Method $1$: Completing the square$f(x) = x^2 + 4x + 4 + 1$$f(x) = (x + 2)^2 + 1$Since $(x + 2)^2 \ge 0$ for all $x \in R$,the minimum value of $f(x)$ is $0 + 1 = 1$.Method $2$: Derivative method$f'(x) = 2x + 4$Setting $f'(x) = 0$ gives $2x + 4 = 0$,so $x = -2$.$f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1$.Thus,the minimum value is $1$.

The minimum value of $f(x) = x^2 + 4x + 5$ is . . . . . . ,where $x \in R$.

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