If you look in the top right-hand corner, it should either say 'rad' or 'deg' next to the battery symbol.

If you click on it with the mouse - it will switch to the other mode. (You will still need to re-do calculations though)
Just be careful - if you press the up/down arrows while the mouse is still over it - it may switch back.

Hope this helps!

2 months later

Hey guys,

How do you find the domain of:
√x-1/x+2 ?

We know that x-1/ x+2 has to be >= 0

I tried doing it in a way that I brought x+2 to the other side and then I had:
x-1 >= 0

But the answer says:
x <-2 U x >=1

How...
Is there a method to do this?

Please help.

Thanks.

    PizzaMaster

    Function:
    √(x-1)/( x+2)

    √(x-1) implies that x≥1 as you cannot have a negative value inside the root (to get a real value). x+2 is the denominator and cannot equal zero. So the domain of x+2 would be any real number except -2. But since the domain of √(x-1) is x≥1, we can disregard the -2 and be left with the final answer of x≥1. I'm not sure if I interpreted your function correctly tho, my answer is correct tho according to my interpretation. Let me know if the function is listed above is the one you are talking about, I CAS verified it as well.

      Thank you for your help.
      However, I meant that the (x-1) and (x+2) are both under the square root sign.
      Sorry for the confusion.

      No worries, in that case we have a whole new situation. The same principle applies in that x+2 cannot equal zero and everything inside the bracket must be 0 or greater. To be honest, I would convert (x-1)/(x+2) into a partial fraction which would yield 1 - 3/(x+2) which can be obtained through long division. If you graph that you get a hyperbola (I think it's called that). You just gotta find when the graph is greater than or equal to zero. I would strongly suggest graphing it and you'll come to realise that from -2<x<1 the graph is negative. Thus the domain of that function would be the direct opposite, aka your aforementioned answer.

      You'll find graphing stuff is always gonna help you and you need to learn and get used to do doing it. Hope that Helps.

      Can a one-to-one function have a many-to one inverse function?
      For example:
      f(x) = 2x^ 3/4 + 1

      has an inverse:
      f-1(x) = [(x-1) / 2]^ 4/3

      How come?

        PizzaMaster
        If you have f(x) = x0.5

        Then f-1(x) = x2, but restricted to the domain R+ u {0}, which means the inverse is also 1-1. The negative branch of x2 isn't a reflection of y = x0.5 in the line y = x, so it's not part of the inverse function

          Ohh so you mean that the range of the original is the domain of the inverse?

          And so that's why
          f(x) = 2x^ 3/4 + 1

          does have an inverse:
          f-1(x) = [(x-1) / 2]^ 4/3
          BUT with domain [1, infinity)

            How do we know whether the question wants us to give the dilation from x axis or y axis when given both the original and transformed function.

            For example:
            f(x)= 1/x2
            f1(x)= 5/x2

            I got dilation by a factor of 1/5 from the y axis but the answers say factor of 5 from the x axis. Is my answer still valid?

              PizzaMaster

              Correct that's why you need to restrict the domain of the original function

                snowflake Your answer wouldn't be considered correct, if the coefficient of x2 were 5 ie y = 1 / (5x2) then it'd be correct since that would be a dilation from the y-axis

                    Billzene

                    Ohh ok but how do we whether to work out it’s dilation from the X or Y axis if it doesn’t say in the question?

                        snowflake by recognition of the form y = a/(1/b(x-h))n + k

                        a = dilation factor from x axis
                        b = dilation factor from y axis
                        h = translation in +ve direction of x axis
                        n = power
                        k = translation in the positive direction of y axis

                        This is similar to turning point form of parabolas

                            5 days later

                            find values of a and b such that a(x+2)+b(x+3)=18x+8 for all values of x
                            if someone could help that would be great

                              chemistry1111 Expand then factorising LHS gives (a+b)x + (2a + 3b) = 18x + 8 --> by comparing coefficients, you obtain a+b = 18 and 2a+3b =8, solve simultaneously for a and b, giving a=46 and b=-28

                                thank you so much that makes sense

                                also this question
                                For the polynomial P(x)=(a+1)x3 + (b-7)x2 +c+5, find values for a,b and c if P(x) has a degree 2, a leading coefficient of 3 and the constant term is -1

                                when factorizing polynomials in methods unit 3/4, what method do most people use vcaa prefer, long division or equating coefficients