H(x), or the range of this function, must be y >= 0, as you can never get a negative when square rooting any number.
For the domain, x cannot exceed -2 or 2, as this would mean we'd be finding the square root of a negative number, which can't happen without getting in to imaginary numbers. Thus, the domain must be -2 <= x <= 2.
Answer:
y = x^(sin(x))
y = e^(SIN(x)*LN(x))
y' = (COS(x)·LN(x) + SIN(x)/x)*e^(SIN(x)*LN(x))
y' = (COS(x)·LN(x) + SIN(x)/x)*x^(SIN(x))
Subtract -6 on both sides
Now you have x/-2 on the left and 3 on the right
Then multiply -2 on both sides, leaving you with x on the left and -6 on the right
So x< -6
Replace x with y and solve for y~
y=2*
x=2^y
log x = y log 2
y=log x / log 2
Hope this helps and leave a brainliest to help me reach expert ;)
Answer:the last box
Step-by-step explanation:the other ones wouldnt make any sense