Functions - Domain and Range

3^x increases as x increases, so put the two ends of the domain into f, and remember what a negative power means.

1. 3^x is increasing, so the smallest output is at x=-2 and the largest at x=2.
2. \mathrm{f}(-2)=3^{-2}=\dfrac{1}{9} (a negative power gives a reciprocal, not a negative number).
3. \mathrm{f}(2)=3^2=9, but x=2 is not in the domain (x<2), so 9 is not reached: use <.
4. x=-2 is included, so use \leqslant at the bottom.
5. The range is written in terms of \mathrm{f}(x), not x.
6. Answer: \dfrac{1}{9}\leqslant \mathrm{f}(x)<9

Complete the square to find the turning point and check whether it lies inside the domain before using the end values.

1. Complete the square: x^2-4x+7=(x-2)^2+3
2. The minimum point is at x=2, which is inside the domain, so the least value is \mathrm{h}(2)=3 (and it is reached, so \leqslant).
3. Check the ends: \mathrm{h}(-1)=1+4+7=12 and \mathrm{h}(3)=9-12+7=4
4. The largest value would be 12, at x=-1, but x=-1 is not in the domain (-1<x), so use <.
5. Answer: 3\leqslant \mathrm{h}(x)<12

The lowest value of a quadratic on a domain is at its turning point if the turning point is inside the domain; the highest value is at one of the ends.

Part (a)

1. Complete the square: x^2-2x+c=(x-1)^2+c-1
2. The minimum point is at x=1, which is inside the domain because -2\leqslant 1\leqslant k (as k>1).
3. So the least value of f is c-1, and this must equal 2.
4. c-1=2
5. Answer: c=3

Part (b)

1. \mathrm{f}(x)=x^2-2x+3
2. At the left end: \mathrm{f}(-2)=4+4+3=11, which is less than 18.
3. So the greatest value, 18, must happen at the other end, x=k.
4. k^2-2k+3=18
5. k^2-2k-15=0
6. (k-5)(k+3)=0, so k=5 or k=-3
7. k>1, so k=5
8. Answer: k=5