Find the extreme values of the function and where they occur.
-$$y = \frac { 6 } { \sqrt { 1 - 3 x ^ { 2 } } }$$
A) The minimum is 0 at x = 1.
B) The maximum is 6 at x = 2.
C) The maximum is 6 at x = -2.
D) The minimum is 6 at x = 0.

Question

Quizplus · Accepted Answer

The function $y = \frac{6}{\sqrt{1 - 3x^2}}$ has a minimum value when the denominator is at its maximum. The denominator, $\sqrt{1 - 3x^2}$, reaches its maximum value when $x = 0$, making the denominator equal to $\sqrt{1}$ or 1. Thus, the minimum value of the function is $y = \frac{6}{1} = 6$ at $x = 0$. There are no maximum values since the function approaches infinity as $x$ approaches $\pm\frac{1}{\sqrt{3}}$.

Find the Extreme Values of the Function and Where They y=61−3x2y = \frac { 6 } { \sqrt { 1 - 3 x ^ { 2 } } }y=1−3x2​6​

Find the Extreme Values of the Function and Where They $y = \frac { 6 } { \sqrt { 1 - 3 x ^ { 2 } } }$