Helpp!! Urgent Mid term practice, i need verification. ASAP

Question

PLEASE HELP OUT!! I really need to confirm my answers. I would really appreciate it

1) Solve. Answer in interval notation, restricting your attention to −𝜋 ≤ x ≤ 𝜋.

sin²(x) < 1/2

I got (0, 𝜋/2) U (𝜋, 3𝜋/2). I think I got this one wrong. Please explain in steps if possible.

2) for the inverse f-1(x)= (9=2x)/(x-1). Use the relationship between the domain and the range of inverse functions to find the range of f-1(x). In interval notation.

I got (-infinity, -2) U (2, positive infinity) ( I don't know if the brackets are right, but I think this is wrong I don't know)

3) The following is a one-to-one function. Find a formula for the inverse function; be mindful of restrictions on the domain of the function and its inverse. (answer should be in terms of x)

y = 10(1 − e^−7x)

I checked Mathway for this as my answer was wrong, they have it as: ln(- ((-x/10) + 1)/ 7)

Answer 1

The range of an inverse function is the same as the domain of the original function. The domain of the original function is all real numbers except 1.

So the range of this inverted function = domain of the original function = (-∞ , 1) ∪ (1 , ∞ ).

Checklist:

Excluded from domain -

Any value that would create a division by zero.

Any value that would lead to a negative even root ( that is, square root, fourth root, sixth root, etc.) if you cannot use complex numbers and are limited to real numbers.

Any value that would lead to a negative logarithm if you cannot use complex numbers and are limited to real numbers.

Answer 2

Here are the answers to (1) and (3); for (2) I could not interpret what you typed well enough to come up with an answer. (You appeared to type an equals sign instead of something else - maybe a "+"?)

(1): If sin² x < 1/2, then sin x must be between -(√2)/2 and (√2)/2. This happens whenever the reference angle for x is between 0 and π/4. Since x must be between -π and π, we have [-π, -3π/4) ∪ (-π/4, π/4) ∪ (3π/4, π].

(3): Mathway is correct. We find the answer by reversing x and y, and then solving the new equation for y in terms of x: x = 10(1 - e^-7y); x/10 = 1 - e^-7y; (x/10) - 1 = -e^-7y; 1 - (x/10) = e^-7y; ln (1 - (x/10)) = -7y; and, finally, y = - (ln (1 - (x/10)))/7. This can be simplified using the relation - ln x = ln (1/x) to y = (ln (10/(10 - x)))/7. That is not exactly what Mathway has, but it is equivalent.

For (2), if that extra equals sign was supposed to be a "+", you are correct and have the bracket notation right as well.