Algebra Know-It-ALL

632 Worked-Out Solutions to Exercises: Chapters 11 to 19

10. When the scale increments in a graph are not the same (as is the case in Fig. 14-13), we
    cannot use the “point reflector” scheme directly to see what the inverse graph looks like.
    It’s better to derive an equation for the inverse relation, and then plot its graph on the
    basis of that equation. We want to derive the inverse of u=t^3. We begin by transposing
    the values of the variables without changing their names, getting

t=u^3

We can take the cube root of both sides of the equation here, and we don’t run any risk

of ambiguity. That’s because the original function is a bijection. Remember what that

means: Every value in the domain has exactly one “mate” in the range, and vice-versa.

There’s no chance for confusion or duplicity as there would be if we were dealing with an

even-numbered power of u. When we take the cube root of both sides, we get

±t1/3=u

Reversing the sense gives us

u=±t1/3

This is the inverse of u=t^3. Figure B-5 is a graph of this relation. Note that the incre-

ments of the scales have been transposed, as well as the points in the graph, so the graph

will fit neatly into the available space.

2

4

6

–6

–40^204060

t

u

(64,4)

(27,3)

(8,2)

(0,0)

(–8,–2)

(–27,–3)

(–64,–4)

Figure B-5 Illustration for the solution to Prob. 10

in Chap. 14.