Question 1208035
<pre>
{{{A(x) = 4x*sqrt(1 - x^2)}}}

We find the x-intercept(s) by setting A(x) = 0

{{{0 = 4x*sqrt(1 - x^2)}}}

{{{4x=0}}};   {{{sqrt(1-x^2)=0}}}
 {{{x=0}}};   {{{1-x^2=0}}}
               {{{1=x^2}}}
               {{{""+-1=x}}}

So the x-intercepts are (-1,0), (0,0), (1,0

We find the y-intercept(s) by setting x = 0  

{{{A(0) = 4(0)*sqrt(1 - (0)^2)}}}
{{{A(0) - 0}}}

So the y-intercept is (0,0) which is both an x- and a y-intercept.


{{{A(x) = 4x*sqrt(1 - x^2)}}}

Since a square root's radicand cannot be negative

{{{1-x^2>=0}}}
{{{(1-x)(1+x)>=0}}}

So either both factor are positive or both negative

{{{matrix(1,3,1-x>=0,AND,1+x>=0)}}} OR {{{matrix(1,3,1-x<=0,AND,1+x<=0)}}}
{{{matrix(1,3,1>=x,AND,x>=-1)}}} OR {{{cross(matrix(1,3,1<=x,AND,x<-1))}}}

So the domain is {{{-1<=x<=1}}} or in interval notation [-1,1]


{{{drawing(400,400,-3,3,-3,3,

red(arc(-4,0,10,-10, 0,8)), red(arc(5,0,12.1,-12.1,180,186)),

graph(400,400,-3,3,-3,3,4x*sqrt(1-x^2)))}}}

For the range, we must find the maxima and minima

{{{A(x) = 4x*sqrt(1 - x^2)}}}

{{{A(x) = 4*(x*(1 - x^2)^("1/2")^"")}}}

{{{dA/dx=4(x*(1/2)(1-x^2)^("-1/2")*(-2x)+(1-x^2)^("1/2")(1)^"")}}}

{{{dA/dx=4(x*(1/cross(2))(1-x^2)^("-1/2")*(-cross(2)x)+(1-x^2)^("1/2")^"")}}}

{{{dA/dx=4(x*(1-x^2)^("-1/2")*(-x)+(1-x^2)^("1/2")^"")}}}

{{{dA/dx=4(-x^2*(1-x^2)^("-1/2")+(1-x^2)^("1/2")^"")}}}

We can factor the -1/2 power out of both terms and since {{{1/2-(-1/2)=1}}},
we will just have the 1st power of the parenthetical expression:

{{{dA/dx= "" +- 4((1-x^2)^("-1/2")(-x^2+(1-x^2)^1)^"")}}}

Now I'll let you simplify that down to

{{{dA/dx="" +- (4-8x^2)/sqrt(1-x^2)}}}

Set that = 0 and you get the x-coordinates of the maximum and minimum points

{{{4-8x^2=0}}}
{{{4=8x^2}}}
{{{4/8=x^2}}}
{{{1/2=x^2}}}
{{{"" +- sqrt(1/2)=x}}}

Substituting:

{{{A("" +- 1/sqrt(2) )}}} = {{{"" +- 4(1/sqrt(2))*sqrt(1 - (1/sqrt(2))^2))}}}

{{{A("" +- 1/sqrt(2) )}}} = {{{("" +- 4/sqrt(2))sqrt(1 - 1/2)}}}

{{{A("" +- 1/sqrt(2) )}}} = {{{"" +- (4/sqrt(2))(1/sqrt(2))}}}

{{{A("" +- 1/sqrt(2) )}}} = {{{"" +- 4/2}}}

{{{A("" +- 1/sqrt(2) )}}} = {{{"" +- 2}}}

So the range is {{{-2<=x<=2}}} or in interval notation [-2,2]

Edwin</pre>