Find the zeros of the function algebraically.? $$f ( x ) = \frac { 1 } { 8 } x ^ { 3 } - x$$ ?
A) $x = 0 , \sqrt { 8 }$
B) $x = 0 , - \sqrt { 8 }$
C) $x = 0 , \pm 8$
D) $x = 0 , \pm \sqrt { 8 }$
E) $x = 0,8$

Question

Quizplus · Accepted Answer

To find the zeros of the function, set $f(x) = 0$:$$0 = \frac{1}{8}x^3 - x$$$$0 = x(\frac{1}{8}x^2 - 1)$$$$0 = x(\frac{x^2}{8} - 1)$$This gives us $x = 0$ and $\frac{x^2}{8} - 1 = 0$, which simplifies to $x^2 = 8$, giving $x = \pm \sqrt{8}$.

Find the Zeros of the Function Algebraically f(x)=18x3−xf ( x ) = \frac { 1 } { 8 } x ^ { 3 } - xf(x)=81​x3−x

Find the Zeros of the Function Algebraically $f ( x ) = \frac { 1 } { 8 } x ^ { 3 } - x$