Find the largest open interval where the function is changing as requested.
-Decreasing $f ( x ) = x ^ { 3 } - 4 x$
A) $\left( \frac { 2 \sqrt { 3 } } { 3 } , \infty \right)$
B) $( - \infty , \infty )$
C) $\left( - \infty , - \frac { 2 \sqrt { 3 } } { 3 } \right)$
D) $\left( - \frac { 2 \sqrt { 3 } } { 3 } , \frac { 2 \sqrt { 3 } } { 3 } \right)$

Question

Quizplus · Accepted Answer

To find the intervals where the function is decreasing, we need to find the critical points where the derivative is zero or undefined, and then test the intervals between those points.

The derivative of the function is:
$f'(x) = 3x^2 - 4$

Setting it to zero, we get:
$3x^2 - 4 = 0$
$3x^2 = 4$
$x^2 = \frac{4}{3}$
$x = \pm \frac{2\sqrt{3}}{3}$

So we have two critical points: $-\frac{2\sqrt{3}}{3}$ and $\frac{2\sqrt{3}}{3}$.

Testing the interval $(-\infty, -\frac{2\sqrt{3}}{3})$, we can pick any number in that interval, say $x=-1$. Then:
$f(-1) = (-1)^3 - 4(-1) = -3$
$f(-0.5) = (-0.5)^3 - 4(-0.5) \approx -3.19$
Since $f(-1) > f(-0.5)$, the function is decreasing in $(-\infty, -\frac{2\sqrt{3}}{3})$.

Testing the interval $(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3})$, we can again pick any number in that interval, say $x=0$. Then:
$f(0) = 0$
$f(0.5) = (0.5)^3 - 4(0.5) \approx -2.69$
Since $f(0) < f(0.5)$, the function is increasing in $(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3})$.

Testing the interval $(\frac{2\sqrt{3}}{3}, \infty)$, we can again pick any number in that interval, say $x=1$. Then:
$f(1) = 1^3 - 4(1) = -3$
$f(2) = 2^3 - 4(2) = 0$
Since $f(1) > f(2)$, the function is decreasing in $(\frac{2\sqrt{3}}{3}, \infty)$.

So the largest open interval where the function is decreasing is $(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3})$. The correct choice is D.

Find the Largest Open Interval Where the Function Is Changing f(x)=x3−4xf ( x ) = x ^ { 3 } - 4 xf(x)=x3−4x

Find the Largest Open Interval Where the Function Is Changing $f ( x ) = x ^ { 3 } - 4 x$