The local maximum for the function $f(x)=8 X_{3}+2 X_{2}-4 X-3$ is obtained when $X$ is equal to
A) 1
B) -1
C) $5 / 6$
D) $1 / 6$

Question

Quizplus · Accepted Answer

The given function is $f(x) = 8x^3 + 2x^2 - 4x - 3$. To find the local maximum, we first find the derivative of the function, which is $f'(x) = 24x^2 + 4x - 4$. Setting the derivative equal to zero to find critical points gives $24x^2 + 4x - 4 = 0$. Solving this quadratic equation, we find that $x = -1$ and $x = 1/6$ are the critical points. To determine which of these points corresponds to a local maximum, we can use the second derivative test or analyze the sign changes of the first derivative around these points. The second derivative of the function is $f''(x) = 48x + 4$. Evaluating $f''(x)$ at $x = -1$ gives a negative value, indicating a concave down graph at this point, which means $x = -1$ corresponds to a local maximum.

The Local Maximum for the Function f(x)=8X3+2X2−4X−3f(x)=8 X_{3}+2 X_{2}-4 X-3f(x)=8X3​+2X2​−4X−3 Is Obtained When

The Local Maximum for the Function $f(x)=8 X_{3}+2 X_{2}-4 X-3$ Is Obtained When