Find all relative minima of the given function. $y = x ^ { 4 } - 8 x ^ { 3 } + 16 x ^ { 2 } + 16$
A) $( 0,16 )$
B) $( 2,32 )$
C) $( 4,16 )$
D) $( 0,16 )$ , $( 4,16 )$
E)no relative minima

Question

Quizplus · Accepted Answer

To find the relative minima, we first find the derivative of the function $y = x^4 - 8x^3 + 16x^2 + 16$, which is $y' = 4x^3 - 24x^2 + 32x$. Setting this derivative equal to zero gives the critical points: $4x^3 - 24x^2 + 32x = 0$, which simplifies to $x(4x^2 - 24x + 32) = 0$, giving $x = 0$ and $x = 4$ (after further factorization or solving the quadratic equation $4x^2 - 24x + 32 = 0$). To determine if these points are minima, we can use the second derivative test or analyze the behavior of the first derivative around these points. The second derivative is $y'' = 12x^2 - 48x + 32$. Plugging in the critical points:- For $x = 0$, $y'' = 32$, which is positive, indicating a local minimum.- For $x = 4$, $y'' = 12(4)^2 - 48(4) + 32 = 192 - 192 + 32 = 32$, which is also positive, indicating another local minimum.Therefore, both $x = 0$ and $x = 4$ are points of relative minima, making the correct answer (D) $(0,16)$ and $(4,16)$.

Find All Relative Minima of the Given Function y=x4−8x3+16x2+16y = x ^ { 4 } - 8 x ^ { 3 } + 16 x ^ { 2 } + 16y=x4−8x3+16x2+16

Find All Relative Minima of the Given Function $y = x ^ { 4 } - 8 x ^ { 3 } + 16 x ^ { 2 } + 16$