Find two linear functions whose domain is $0 \leqslant t \leqslant 2$ and whose range is $1 \leqslant f(t) \leqslant 9$

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I'm supposed to find two linear functions for the same domain and range, I have found one. How do I find the other?

The one I found:

Since Domain is $0 \leqslant t \leqslant 2$And Range is $1 \leqslant f(t) \leqslant 9$

I took $f(0) = 1$ and considered constant $C = 1$

Then I took $f(2) = 9$ , Since linear equations are of the form $f(t) = vt + C$, v being slope and C being constant.

$$f(2) = v(2) + 1\implies 9 = v(2) + 1\implies v = 4$$

Thus the linear equation becomes $f(t) = 4t + 1$

How do I find the second linear function with the same Domain and Range?

This is the $24$th question in Strang's Calculus at section $1.1$

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3 Answers

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Flip it: consider the case where $f(0) = 9$ and $f(2) = 1$ instead. You shall obtain:

$$f(t) = 9-4t$$

Graphs to illustrate the flip:

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For the function you found, $f(0)=1$ and $f(2)=9$. It is quite clear from a geometric perspective that the other linear function has $g(0)=9$ and $g(2)=1$.

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Once you get $f(t)=4t+1$, notice that $$1 \le f(t) \le 9 \iff 1 \le 10-f(t) \le 9$$ so $g(t)=10-f(t)=9-4t$ is the other linear function.

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