Question

what does mean that the function 1+3sin^4(2x) is going to be bounded since the sine is never larger than 1 ?

Answer 1

does it mean that at the roots value the function is unbounded?

Answer 2

Since \[|\sin(x)| \leq 1\]You know that \[|3\;\sin^4(2x)|\leq 3\]Thus, \[1+3\;\sin^4(2x) \leq4\]Likewise, you also know that \[1+3\;\sin^4(2x) \geq -2\]Since the range of this function is bounded within the interval \([-2, 4]\), the function is bounded.

Answer 3

how did you find the range so quickly?

Answer 4

yes, where did you get -2?

Answer 5

did you use this tirg identity sin^2(2x) - 1/2(1-cos(2x)) ?

Answer 6

sin^2(2x) = 1/2(1-cos(2x))

Answer 7

I got the -2 from \[|3\;\sin^4(2x)|\leq3\Rightarrow -3\leq 3\;\sin^4(2x)\Rightarrow -2\leq1+3\;\sin^4(2x)\]However, I realize now that you can restrict it to \(+1\) instead of \(-2\).

Answer 8

i understand that 3 is the amplitude, and then you add one to get minimum of -2, but wouldnt the exponent have an effect?

Answer 9

yeah, i got 1

Answer 10

@mridrik you are correct. Because you have a \(\sin^4(x)\), it must always be positive, since any negative number raised to an even power will be positive.

Answer 11

thanks @ all

Answer 12

You're welcome.

Answer 13

this website is really awesome because I am self teaching myself & whenever I am stuck and google doesn't help then someone later or sooner resolves my problem

Answer 14

yeah, i agree, kindgeorge, what would the range be if sin was taken to the 3rd power instead?

Answer 15

kinggeorge*

Answer 16

If it was to the third, it would be at -2 instead of +1. If the 3 was contained in the power it would be -26.

Answer 17

? -26? where'd you get that?

Answer 18

\((-3)^3=-27\) so if you add one you get \(-26\)

Answer 19

so the minimum is -26?

Answer 20

what is the proceadure for finding the range? (especially in this case)

Answer 21

If the function were instead \[1+(3\sin(2x))^3\]the minimum would be \(-26\)

Answer 22

i meant 3sin^3, but continue

Answer 23

To find the range (in this case), first find the range of \(\sin(x)\) This is \([-1, 1]\)Then, if you have an even exponent, the range become \([0, 1]\). If you have an odd exponent, nothing happens.

Once you have that, multiply the range by any number in front of the sin. In this case you have a 3, so the range goes to \([0*3, 1*3]=[0, 3]\). Finally, add/subtract any constant in the front so in this case it becomes \([1, 4]\).

Answer 24

Well explained!

Answer 25

@mridrik the range of \[1+3\;\sin^3(2x)\]would be \([-2, 4]\)

Answer 26

right!

Answer 27

what is the method to find the domain?

Answer 28

To find the domain, just find the numbers the function is defined on. For this function, the domain would be \((-\infty, \infty)\) because all the real numbers are defined for \(2x\), \(\sin(x)\), \(x^4\), \(3x\) and \(1+x\) (Your function uses all of these functions).

If one of these wasn't defined for all numbers, then the domain of your function would not include those numbers it's being divided by. If, for example, your function was \[1+{3\;\sin^4 (2x)\over x}\]Then the domain would be \((-\infty, 0), (0, \infty)\) because \(x=0\) isn't defined for \(1\over x\).

Answer 29

thankssssssssssss

Answer 30

You're welcome.