Question

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

Answer 1

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

Answer 2

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]

Answer 3

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

Answer 4

let me check

Answer 5

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]

Answer 6

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

Answer 7

oh okay makes sense :) thank you :D

Answer 8

welcome, sorry i made a mistake earlier

Answer 9

np :) thank you.