Use a computer algebra system to graph the function $\frac { 4 } { 3 } x \sqrt { 3 - x } \text { and determine all }$ absolute extrema on the interval $[ 0,3 ]$
A) absolute maximum: $\left( 2 , \frac { 8 } { 3 } \right)$ absolute minimum: $( 0,0 ) ( 3,0 )$
B) absolute maximum: $\left( 1 , \frac { 4 } { 3 } \right)$ absolute minimum: $( 0,0 ) ( 3,0 )$
C) absolute maximum: $( 3,4 )$ absolute minimum: $( 0,0 )$
D) absolute maximum: $( 3,4 )$ absolute minimum: $( 0,3 )$
E) absolute maximum: $\left( 2 , \frac { 8 } { 3 } \right)$ absolute minimum: $( 0,3 )$

Question

Use a computer algebra system to graph the function $\frac { 4 } { 3 } x \sqrt { 3 - x } \text { and determine all }$ absolute extrema on the interval $[ 0,3 ]$ 
A) absolute maximum: $\left( 2 , \frac { 8 } { 3 } \right)$ absolute minimum: $( 0,0 ) ( 3,0 )$
B) absolute maximum: $\left( 1 , \frac { 4 } { 3 } \right)$ absolute minimum: $( 0,0 ) ( 3,0 )$
C) absolute maximum: $( 3,4 )$ absolute minimum: $( 0,0 )$
D) absolute maximum: $( 3,4 )$ absolute minimum: $( 0,3 )$
E) absolute maximum: $\left( 2 , \frac { 8 } { 3 } \right)$ absolute minimum: $( 0,3 )$

Quizplus · Accepted Answer

Using a computer algebra system (such as Desmos), we can graph the function and find that there is a relative maximum at $x = 2$ and two endpoints at $x = 0$ and $x = 3$. We can then check the function values at these three points and at any critical points to determine the absolute extrema. Taking the derivative of the function and setting it equal to zero, we find that the critical point is at $x = \frac{3}{4}$, but this point is not in the interval $[0,3]$. Thus, we only need to check the function values at the endpoints and at the relative maximum:
$$\text{Function value at } x = 0: \frac{4}{3}(0)\sqrt{3-0} = 0$$
$$\text{Function value at } x = 2: \frac{4}{3}(2)\sqrt{3-2} = \frac{8}{3}$$
$$\text{Function value at } x = 3: \frac{4}{3}(3)\sqrt{3-3} = 0$$
Therefore, the absolute minimum is at both endpoints $(0,0)$ and $(3,0)$, and the absolute maximum is at the relative maximum $\left(2,\frac{8}{3}\right)$. Thus, the answer is \textbf{A}.

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