## anonymous 4 years ago y=(sinx)(sqrt{25-x^2}), what is the range? how do find that?

1. anonymous

$y=(sinx)(\sqrt{25-x^2})$

2. ash2326

let's find out the domain first sin x --x can be any real number sqrt (25-x^2) 25-x^2>=0 or x^2<=25 x will be in the [-5,+5] so the domain is [-5, 5] max value of square root will occur when x=0, it'll be 5 now sin x' s range is [-1,1] so the range of the function is [-5,5]

3. anonymous

oh okay, thank you. but it says estimate the range to two decimal places, explain why only an estimate is possible...

4. ash2326

let me check

5. ash2326

sorry i made a mistake, for the max value when x=0 root 25=5 but sin 0 =0 sin is max when x=pi/2 i.e when x=1.57 at x=1.57 the value of y is 4.74711 sin x is minimum at x=-pi/2 or -1.57 therefore the minimum value is -4.74711 range is from [-4.74711, 4.74711]

6. anonymous

oh okay, so you basically plug in 1.57 and -1.57 into the function to get the y-value?

7. anonymous

oh okay makes sense :) thank you :D

8. ash2326

welcome, sorry i made a mistake earlier

9. anonymous

np :) thank you.