12 multiple-choice questions, progressively harder.
Why can we speak of the inverse of fff rather than merely an inverse?
Solution
Correct answer: C
If ggg and hhh both invert fff, associativity collapses them.
g=g∘(f∘h)=(g∘f)∘h=hg = g \circ (f \circ h) = (g \circ f) \circ h = hg=g∘(f∘h)=(g∘f)∘h=h
One-to-oneness gives existence of an inverse, but uniqueness is what associativity provides, and that is what licenses the word 'the'.
For f(x)=−x3f(x) = -x^3f(x)=−x3 and its inverse f−1(x)=−x3f^{-1}(x) = -\sqrt[3]{x}f−1(x)=−3x, at which point do the two graphs meet OFF the line y=xy = xy=x?
Correct answer: D
Set f(x)=f−1(x)f(x) = f^{-1}(x)f(x)=f−1(x): from −x3=−x3-x^3 = -\sqrt[3]{x}−x3=−3x you get x∈{−1,0,1}x \in \{-1, 0, 1\}x∈{−1,0,1}.
f(1)=−1andf−1(1)=−13=−1f(1) = -1 \quad\text{and}\quad f^{-1}(1) = -\sqrt[3]{1} = -1f(1)=−1andf−1(1)=−31=−1
So (1,−1)(1, -1)(1,−1) is on both graphs and off y=xy = xy=x. The point (0,0)(0, 0)(0,0) is a meeting point but lies on y=xy = xy=x; (2,2)(2, 2)(2,2) and (−2,−2)(-2, -2)(−2,−2) are on y=xy = xy=x yet are not on the curves at all.
Let f(x)=x2f(x) = x^2f(x)=x2 with codomain all real numbers (not just its range). Is fff restricted to x≥0x \ge 0x≥0 invertible as a map to this codomain?
Correct answer: B
On x≥0x \ge 0x≥0 the rule is one-to-one, but a two-sided inverse to R\mathbb{R}R needs every real to be an output.
f(g(y))=y is impossible for y<0, since x2≥0f(g(y)) = y \text{ is impossible for } y < 0, \text{ since } x^2 \ge 0f(g(y))=y is impossible for y<0, since x2≥0
The missing negative outputs block f∘g=idf \circ g = \operatorname{id}f∘g=id, so with this larger codomain fff is not invertible. It becomes invertible only once the codomain is cut down to its range.
Which computation correctly verifies that g(x)=x−12g(x) = \dfrac{x - 1}{2}g(x)=2x−1 is the inverse of f(x)=2x+1f(x) = 2x + 1f(x)=2x+1?
Correct answer: A
Verification means composing both ways and getting xxx each time.
g(f(x))=(2x+1)−12=x,f(g(x))=2⋅x−12+1=xg(f(x)) = \frac{(2x + 1) - 1}{2} = x, \qquad f(g(x)) = 2 \cdot \frac{x - 1}{2} + 1 = xg(f(x))=2(2x+1)−1=x,f(g(x))=2⋅2x−1+1=x
A product equal to 111 or a single-sided check does not establish an inverse.
Suppose g∘f=idg \circ f = \operatorname{id}g∘f=id on domf\operatorname{dom} fdomf but you have not checked f∘gf \circ gf∘g. What can you conclude?
Apply ggg to a collision: if f(a)=f(b)f(a) = f(b)f(a)=f(b) then a=g(f(a))=g(f(b))=ba = g(f(a)) = g(f(b)) = ba=g(f(a))=g(f(b))=b, so fff is one-to-one.
g∘f=id⇒f one-to-one, but f∘g may failg \circ f = \operatorname{id} \Rightarrow f \text{ one-to-one, but } f \circ g \text{ may fail}g∘f=id⇒f one-to-one, but f∘g may fail
On an infinite domain a left inverse need not be a right inverse, so ggg is not yet confirmed as the inverse.
The function f(x)=x2f(x) = x^2f(x)=x2 can be given an inverse by restricting its domain. How many different inverses arise from the restrictions x≥0x \ge 0x≥0 and x≤0x \le 0x≤0?
Each restriction is a genuinely different function, and each has its own inverse.
x≥0⇒f−1(x)=x,x≤0⇒f−1(x)=−xx \ge 0 \Rightarrow f^{-1}(x) = \sqrt{x}, \qquad x \le 0 \Rightarrow f^{-1}(x) = -\sqrt{x}x≥0⇒f−1(x)=x,x≤0⇒f−1(x)=−x
So two different inverses arise; there is no single 'the' restriction.
A one-to-one function fff has f(0)=1f(0) = 1f(0)=1, f(1)=3f(1) = 3f(1)=3, and f(2)=0f(2) = 0f(2)=0. What is f−1(0)f^{-1}(0)f−1(0)?
Find the input whose output is 000.
f(2)=0 ⇒ f−1(0)=2f(2) = 0 \;\Rightarrow\; f^{-1}(0) = 2f(2)=0⇒f−1(0)=2
The inverse reads the table backward, output to input.
f(x)=x3+xf(x) = x^3 + xf(x)=x3+x. Is fff invertible on R\mathbb{R}R, as a map onto its range?
If a<ba < ba<b then a3<b3a^3 < b^3a3<b3 (from b3−a3=(b−a)(a2+ab+b2)>0b^3 - a^3 = (b - a)(a^2 + ab + b^2) > 0b3−a3=(b−a)(a2+ab+b2)>0) and a<ba < ba<b, so adding gives a3+a<b3+ba^3 + a < b^3 + ba3+a<b3+b.
a<b ⇒ f(a)<f(b)a < b \;\Rightarrow\; f(a) < f(b)a<b⇒f(a)<f(b)
So fff is strictly increasing, hence one-to-one and invertible onto its range. Lacking a tidy inverse formula does not affect existence.
You are told g(f(x))=xg(f(x)) = xg(f(x))=x for all x∈domfx \in \operatorname{dom} fx∈domf AND f(g(x))=xf(g(x)) = xf(g(x))=x for all x∈domgx \in \operatorname{dom} gx∈domg. What follows?
Both identities are exactly the two-sided definition of an inverse.
g∘f=id and f∘g=id ⇒ g=f−1g \circ f = \operatorname{id} \text{ and } f \circ g = \operatorname{id} \;\Rightarrow\; g = f^{-1}g∘f=id and f∘g=id⇒g=f−1
Since the inverse is unique, ggg is the inverse; no further check is required.
f(x)=x−2f(x) = \sqrt{x - 2}f(x)=x−2 has domain x≥2x \ge 2x≥2 and range y≥0y \ge 0y≥0. What is f−1f^{-1}f−1 and its domain?
Solve y=x−2y = \sqrt{x - 2}y=x−2, then read off the domain as the range of fff.
y2=x−2 ⇒ x=y2+2,domf−1=rangef=[0,∞)y^2 = x - 2 \;\Rightarrow\; x = y^2 + 2, \qquad \operatorname{dom} f^{-1} = \operatorname{range} f = [0, \infty)y2=x−2⇒x=y2+2,domf−1=rangef=[0,∞)
So f−1(x)=x2+2f^{-1}(x) = x^2 + 2f−1(x)=x2+2 on x≥0x \ge 0x≥0; extending it to all reals would invent inputs the inverse cannot take.
The graph of a one-to-one function fff lies entirely in the second quadrant (where x<0x < 0x<0 and y>0y > 0y>0). Where does the graph of f−1f^{-1}f−1 lie?
Reflecting across y=xy = xy=x swaps coordinates.
(x,y)↦(y,x),x<0, y>0 ⇒ new point has first coord>0, second<0(x, y) \mapsto (y, x), \qquad x < 0,\ y > 0 \;\Rightarrow\; \text{new point has first coord} > 0,\ \text{second} < 0(x,y)↦(y,x),x<0, y>0⇒new point has first coord>0, second<0
A point with x>0x > 0x>0 and y<0y < 0y<0 is in the fourth quadrant.
f(x)=x5+x3+xf(x) = x^5 + x^3 + xf(x)=x5+x3+x is strictly increasing, hence one-to-one. Given f(1)=3f(1) = 3f(1)=3, what is f−1(3)f^{-1}(3)f−1(3)?
The inverse returns the input that produced the output.
f(1)=1+1+1=3 ⇒ f−1(3)=1f(1) = 1 + 1 + 1 = 3 \;\Rightarrow\; f^{-1}(3) = 1f(1)=1+1+1=3⇒f−1(3)=1
Because fff is one-to-one, this preimage is unique, so the value is determined.
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