hmm. We have
f(x) = x/2 √(4-x^2) + 2arcsin(x/2)
now, the domain of f(x) is [-2,2],
Now, you might think that the range would thus be the sum of the ranges of
x/2 √(4-x^2): [-1,1]
2arcsin(x/2): [-π,π]
But 2arcsin(x/2) is strictly increasing on [-2,2]
while x/2 √(4-x^2) dips from 0 to -1 and back to 0, and then rises to 1 and drops back to 0 again. So, it does not extend the range of f(x), which is ultimately just [-π,π]
Now, that's not one of the choices. Why not? What assumption must be made?
Find the range of the function F(x)=integral from 0 to x of √(4-t^2) dt
a) [0,4π]
b) [0,π]
c) [-4,0]
d) [0,4]
3 answers
int (2^2-t^2)^.5 dt = (1/2)[t sqrt(4-t^2) +4 sin^-1(t/2) ]
at t = 0 that is 0
at t = x
(1/2)[x sqrt(4-x^2) +4 sin^-1(x/2) ]
= (x/2) sqrt(4-x^2) + 2 sin^-1 (x/2)
well, |x| better be less than 2 or that sqrt is imaginary
If x = 0, it is 0
if x = 2, it is 2 pi/2 or pi so I would pick 0 to pi
when x = -2pi
-pi
at t = 0 that is 0
at t = x
(1/2)[x sqrt(4-x^2) +4 sin^-1(x/2) ]
= (x/2) sqrt(4-x^2) + 2 sin^-1 (x/2)
well, |x| better be less than 2 or that sqrt is imaginary
If x = 0, it is 0
if x = 2, it is 2 pi/2 or pi so I would pick 0 to pi
when x = -2pi
-pi
Thank you both! I worked it out and also got π for when x=2 and -π for when x=-2 and so since that isn't an answer choice I'm wondering if it is 0 to π because the integral is from 0 to x?? Is that the assumption you're referring to?
2 answers