Find the exact location of all the relative and absolute extrema of the function with domain $( - \infty , \infty )$ . $g ( x ) = \frac { 1 } { 4 } x ^ { 4 } - \frac { 2 } { 3 } x ^ { 3 } + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 2 }$
A) $\left( 0 , \frac { 1 } { 5 } \right)$ - absolute maximum
B) $\left( 0 , \frac { 1 } { 5 } \right)$ - absolute minimum
C) $\left( 0 , \frac { 1 } { 3 } \right)$ - relative maximum
D) $\left( 0 , \frac { 1 } { 2 } \right)$ - absolute minimum
E) $( 0,1 )$ - relative maximum

Question

Find the exact location of all the relative and absolute extrema of the function with domain $( - \infty , \infty )$ .   $g ( x ) = \frac { 1 } { 4 } x ^ { 4 } - \frac { 2 } { 3 } x ^ { 3 } + \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 2 }$  
A) $\left( 0 , \frac { 1 } { 5 } \right)$ - absolute maximum
B) $\left( 0 , \frac { 1 } { 5 } \right)$ - absolute minimum
C) $\left( 0 , \frac { 1 } { 3 } \right)$ - relative maximum
D) $\left( 0 , \frac { 1 } { 2 } \right)$ - absolute minimum
E) $( 0,1 )$ - relative maximum

Quizplus · Accepted Answer

To find the extrema of $g(x) = \frac{1}{4}x^4 - \frac{2}{3}x^3 + \frac{1}{2}x^2 + \frac{1}{2}$, we first find its derivative: $g'(x) = x^3 - 2x^2 + x$. Setting $g'(x) = 0$ gives the critical points $x = 0, 1$. Evaluating $g(x)$ at these points and at the endpoints of the domain (which are $-\infty$ and $+\infty$, but for a polynomial of even degree with a positive leading coefficient, the endpoints go to $+\infty$), we find that $g(0) = \frac{1}{2}$ is the lowest value on the entire domain, making it an absolute minimum. Choices A, B, C, and E do not correctly identify the nature and location of the extrema based on the given function.

Find the Exact Location of All the Relative and Absolute (−∞,∞)( - \infty , \infty )(−∞,∞)

Find the Exact Location of All the Relative and Absolute $( - \infty , \infty )$