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<u>Given function</u>
- h(x) = 2x - 1
#15 Find the inverse of h(x)
<u>Substitute x with y and h(x) with x and solve for y:</u>
- x = 2y - 1
- 2y = x + 1
- y = 1/2x + 1/2
<u>The inverse is:</u>
- h⁻¹(x) = 1/2x + 1/2
#16 The graph with both lines is attached.
The x- and y-intercepts of both functions have reversed values.
#17 Table of the inverse function will contain same numbers with swapped domain and range.
<u>Initial look is like this:</u>
- <u>x | -3 | -2 | -1 | 0 | 1 | 2 | 3</u>
- h⁻¹(x) | -1 | | 0 | | 1 | | 2
<u>The rest of the table is filled in by finding the values:</u>
- <u>x | -3 | -2 | -1 | 0 | 1 | 2 | 3</u>
- h⁻¹(x) | -1 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2
The problem is:
We now do L.C.M. of numerator and add, also do L.C.M. of denominator and add together:
Dividing by a fraction is same as multiplying by its reciprocal, so we have:
x cancels out, so can multiply and write:
This is the simplified form.
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