Question

the function $f$ is defined by the given equation. for what value of $x$ does $f(x)$ reach its minimum?
answer preview:

Explanation:

Step1: Analyze the square - term property

Any real - number squared is non - negative. That is, for any real number \(a\), \(a^{2}\geq0\). In the function \(f(x)=(x + 7)^{2}+4\), let \(a=x + 7\), so \((x + 7)^{2}\geq0\).

Step2: Find the minimum of the function

To minimize \(f(x)=(x + 7)^{2}+4\), we need to minimize the term \((x + 7)^{2}\). Since \((x + 7)^{2}\geq0\), the minimum value of \((x + 7)^{2}\) occurs when \((x + 7)^{2}=0\).

Step3: Solve for \(x\)

Set \((x + 7)^{2}=0\). Taking the square root of both sides, we get \(x+7 = 0\). Solving for \(x\), we have \(x=-7\).

Answer:

\(-7\)