The local minimum for the function $f(X)=8 X_{2}-4 X-3$ occurs at $X$ equal to
A) 2
B) 5
C) 0.25
D) 4

Question

Quizplus · Accepted Answer

To find the local minimum of the function $f(X)=8X_{2}-4X-3$, we first need to correct the function's notation, assuming it's meant to be $f(X) = 8X^2 - 4X - 3$. We differentiate the function with respect to $X$ to find the critical points: $f'(X) = 16X - 4$. Setting the derivative equal to zero gives $16X - 4 = 0$, solving for $X$ gives $X = 0.25$. This is the critical point, and by examining the second derivative, $f''(X) = 16$, which is positive, indicating a local minimum at $X = 0.25$.

The Local Minimum for the Function f(X)=8X2−4X−3f(X)=8 X_{2}-4 X-3f(X)=8X2​−4X−3 Occurs At XXX

The Local Minimum for the Function $f(X)=8 X_{2}-4 X-3$ Occurs At $X$