How do you find the domain and range of sqrt(25-(x-2)^2) +325(x2)2+3?

1 Answer
Jan 30, 2017

The domain is x in [-3,7]x[3,7]
The range is y in [3,8]y[3,8]

Explanation:

Let g(x)=sqrt(25-(x-2)^2)+3g(x)=25(x2)2+3

What is under the sqrt sign is >=00. this is the domain

So,

25-(x-2)^2>=025(x2)20

25-(x^2-4x+4)>=025(x24x+4)0

x^2-4x+4-25<=0x24x+4250

x^2-4x-21<=0x24x210

Let's factorise

(x-7)(x+3)<=0(x7)(x+3)0

Let f(x)=(x-7)(x+3)f(x)=(x7)(x+3)

Let 's do a sign chart to solve this inequality

color(white)(aaaa)aaaaxxcolor(white)(aaaa)aaaa-oocolor(white)(aaaa)aaaa-33color(white)(aaaa)aaaa77color(white)(aaaa)aaaa+oo+

color(white)(aaaa)aaaax+3x+3color(white)(aaaaa)aaaaa-color(white)(aaaa)aaaa++color(white)(aaaa)aaaa-

color(white)(aaaa)aaaax-7x7color(white)(aaaaa)aaaaa-color(white)(aaaa)aaaa-color(white)(aaaa)aaaa-

color(white)(aaaa)aaaaf(x)f(x)color(white)(aaaaaa)aaaaaa++color(white)(aaaa)aaaa-color(white)(aaaa)aaaa++

Therefore,

f(x)<=0f(x)0 when x in [-3,7]x[3,7], this is the domain

To calculate the range,

When x=-3x=3, =>, g(-3)=3g(3)=3

When x=7x=7, =>, g(7)=3g(7)=3

When x=2x=2, =>, g(2)=8g(2)=8

Let y=sqrt(25-(x-2)^2)+3y=25(x2)2+3

The range is y in [3,8]y[3,8]

graph{(sqrt(25-(x-2)^2)+3) [-9.74, 12.76, -2.055, 9.195]}