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# stuck on another function question watch

1. im useless at these
2. a) the inverse function is 3 - e^(x/2)

b) put -4 for x in the inverse function

c) i) the first step is to sketch ln x
the second step is to sketch ln ( x - 3 ) which is a translation to the right of 3 units
the third step is to sketch ln -(x-3) which is a reflection in the y axis
the fourth step is to apply a vertical stretch of scale factor 3

ii) reflect any parts of the previous graph which are below the x axis so they are now above the x axis
3. a) ....the domain of the inverse function must be indentical to the range of the original function. the range of f(x) appears to be all real values
5. (Original post by the bear)
a) the inverse function is 3 - e^(x/2)

b) put -4 for x in the inverse function

c) i) the first step is to sketch ln x
the second step is to sketch ln ( x - 3 ) which is a translation to the right of 3 units
the third step is to sketch ln -(x-3) which is a reflection in the y axis
the fourth step is to apply a vertical stretch of scale factor 3

ii) reflect any parts of the previous graph which are below the x axis so they are now above the x axis

could you go through how you got the inverse function as i dont quite understand
thanks
6. (Original post by lil_kk)
could you go through how you got the inverse function as i dont quite understand
thanks
f(x) = 2ln(3-x)

Let y = f(x). Doing this, and then rearranging to find x in terms of y will give you the solution. You then jut replace y with x.

y = 2ln(3-x)
½y = ln(3-x)
e½y = 3-x
x = 3 - e½y

Hence f-1(x) = 3 - e½x
7. (Original post by JohnSPals)
f(x) = 2ln(3-x)

Let y = f(x). Doing this, and then rearranging to find x in terms of y will give you the solution. You then jut replace y with x.

y = 2ln(3-x)
½y = ln(3-x)
e½y = 3-x
x = 3 - e½y

Hence f-1(x) = 3 - e½x
thanks for the explanation
much appreciated

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