I need to find the domain of the following function:
y=arcsin((x-3)/(2x+1))
My math is a bit rusty after holidays...This is what I did: (I'll use <> for the sign of 'not equal')
-> 2x+1<>0 <=> 2x<>-1 <=> x <> -1/2
But i'm not sure about the rest! What I though of doing was this: the domain of arcsin(x) is [-1,1], so I could write that for the given function, ((x-3)/(2x+1)) is between -1 and 1... Not sure though...Help!
To find the domain of the function y = arcsin((x-3)/(2x+1)), there are a few steps you can follow:
Step 1: Identify any values of x that make the denominator, 2x + 1, equal to zero. In this case, you correctly found x ≠ -1/2.
Step 2: Next, consider the range of the arcsin function, which is defined as [-1, 1]. This means that the expression (x-3)/(2x+1) must also fall within this range for the function to be defined.
Step 3: To find the potential values of x that satisfy (x-3)/(2x+1) ≤ 1, solve this inequality. Start by multiplying both sides of the inequality by (2x+1) to eliminate the denominator:
(x-3) ≤ (2x+1)
Step 4: Simplify and solve for x:
x - 3 ≤ 2x + 1
-x ≤ 4
x ≥ -4
In this step, notice that we flipped the inequality sign when dividing by -1.
Step 5: Repeat the process for the inequality (x-3)/(2x+1) ≥ -1:
(x-3) ≥ - (2x+1)
x - 3 ≥ -2x - 1
3x ≥ 2
x ≥ 2/3
Step 6: Finally, combine the results from Step 4 and Step 5 to find the intersection of the solution sets:
x ≥ 2/3 and x ≥ -4
Since the common solution for these two inequalities is x ≥ 2/3, the domain of the function y = arcsin((x-3)/(2x+1)) is x ≥ 2/3.