f(x)=x^2 - 48x + 2,304 what is the minimum value of the given function? answer preview

Explanation:

Step1: Find the derivative

The function is $f(x)=x^{2}-48x + 2304$. Using the power - rule $(x^n)'=nx^{n - 1}$, the derivative $f'(x)=2x-48$.

Step2: Set the derivative equal to zero

Set $f'(x)=0$, so $2x - 48=0$. Solving for $x$ gives $2x=48$, then $x = 24$.

Step3: Determine if it's a minimum

The second - derivative $f''(x)=(2x - 48)'=2>0$. Since $f''(x)>0$, $x = 24$ is a point of minimum.

Step4: Find the minimum value

Substitute $x = 24$ into the original function $f(x)$. $f(24)=24^{2}-48\times24 + 2304=576-1152 + 2304=1728$.

Answer:

$1728$