Question

Locate the absolute extrema of the function over each interval. f(x)=sqrt(4-x^2)
a) [-2,2] b) [-2,0) c) (-2,2) d)[1,2)
I'd appreciate if you could please explain each step and explain how the parenthesis and brackets differ and result in different answers.

Answer 1

before you start, do you know what the graph of \(y=\sqrt{4-x^2}\) is ?

Answer 2

a parabola?

Answer 3

no
ok we start here
the equation
\[x^2+y^2=4\] is a circle centered at the origin with radius 2
if you solve for \(y\) you get
\[y=\pm\sqrt{4-x^2}\] and if you ignore the minus part, you get
\[y=\sqrt{4-x^2}\] which is the upper half semicircle

Answer 4

therefore the extrema on \([-2,2]\)is as follows:
minimum is 0 when \(x=2\) or \(x=-2\)
maximum is 2 when \(x=0\)

Answer 5

on \([-2,0)\) there the minimum is 0 at \(x=-2\) and the max doesn't exist because your interval does not contain the number 0, so as x gets closer to 0, y gets closer to 2, but the 2 is never achieved

Answer 6

on \((-2,2)\) the max is 2 when \(x=0\) but the minimum doesn't exist, because neither \(-2\) nor \(2\) are in your interval. again the function approaches 0 at both endpoints, but because the endpoints are not in the interval, there is no minimum

Answer 7

and finally as \(f\) is decreasing on \((0,2)\) the max on \([1,2)\) will be \(f(1)=\sqrt{3}\) but the min does not exist

Answer 8

Thanks a bunch :)

Answer 9

yw