Find all relative maxima of the given function. $y = x ^ { 4 } - 8 x ^ { 3 } + 16 x ^ { 2 } + 5$
A) $( 0,5 )$
B) $( 2,21 )$
C) $( 4,5 )$
D) $( 0,5 )$ , $( 4,5 )$
E)no relative maxima

Question

Quizplus · Accepted Answer

To find the relative maxima of the function $y = x^4 - 8x^3 + 16x^2 + 5$, we first find its derivative: $y' = 4x^3 - 24x^2 + 32x$. Setting $y'$ equal to zero gives the critical points: $4x^3 - 24x^2 + 32x = 0$, which simplifies to $x(x^2 - 6x + 8) = 0$, yielding $x = 0$, $x = 2$, and $x = 4$. To determine which of these points are maxima, we examine the second derivative, $y'' = 12x^2 - 48x + 32$, and evaluate it at each critical point. At $x = 2$, $y'' < 0$, indicating a relative maximum. Therefore, the relative maximum occurs at $x = 2$, and substituting $x = 2$ into the original function gives $y = 2^4 - 8(2)^3 + 16(2)^2 + 5 = 21$, making the point $(2,21)$ a relative maximum.

Find All Relative Maxima of the Given Function y=x4−8x3+16x2+5y = x ^ { 4 } - 8 x ^ { 3 } + 16 x ^ { 2 } + 5y=x4−8x3+16x2+5

Find All Relative Maxima of the Given Function $y = x ^ { 4 } - 8 x ^ { 3 } + 16 x ^ { 2 } + 5$