12 multiple-choice questions, progressively harder.
The graph of a one-to-one function fff passes through (1,2)(1, 2)(1,2), (2,5)(2, 5)(2,5), and (5,8)(5, 8)(5,8). Using the graph of f−1f^{-1}f−1, evaluate f−1(5)f^{-1}(5)f−1(5).
Solution
Correct answer: D
Find the point on fff whose output is 555, then swap its coordinates. Since (2,5)(2, 5)(2,5) is on fff, its reflection (5,2)(5, 2)(5,2) is on f−1f^{-1}f−1.
f−1(5)=2f^{-1}(5) = 2f−1(5)=2
The inverse reads the input that produced the output 555.
A function fff has inverse f−1(x)=2x+6f^{-1}(x) = 2x + 6f−1(x)=2x+6. What is f(x)f(x)f(x)?
Correct answer: A
The inverse of the inverse is fff itself, so invert f−1(x)=2x+6f^{-1}(x) = 2x + 6f−1(x)=2x+6 by swapping and solving. From y=2x+6y = 2x + 6y=2x+6, swap to x=2y+6x = 2y + 6x=2y+6.
x−6=2y ⇒ y=x−62x - 6 = 2y \;\Rightarrow\; y = \frac{x - 6}{2}x−6=2y⇒y=2x−6
So f(x)=x−62f(x) = \dfrac{x - 6}{2}f(x)=2x−6.
The point (k,3)(k, 3)(k,3) is on the graph of fff. For which value of kkk is this point a fixed point of the reflection across y=xy = xy=x (its own mirror image)?
Correct answer: B
A point is fixed by the reflection when it equals its coordinate-swap, that is, when (k,3)=(3,k)(k, 3) = (3, k)(k,3)=(3,k).
k=3k = 3k=3
Then the point is (3,3)(3, 3)(3,3), which lies on the line y=xy = xy=x.
g(x)=x−2g(x) = \sqrt{x - 2}g(x)=x−2 has domain x≥2x \ge 2x≥2 and range y≥0y \ge 0y≥0. What is the domain of g−1g^{-1}g−1?
Correct answer: C
The domain of the inverse is the range of ggg.
domain of g−1=range of g=x≥0\text{domain of } g^{-1} = \text{range of } g = x \ge 0domain of g−1=range of g=x≥0
Reflecting across y=xy = xy=x turns the outputs y≥0y \ge 0y≥0 into the allowed inputs of g−1g^{-1}g−1.
For g(x)=x−2g(x) = \sqrt{x - 2}g(x)=x−2 (with x≥2x \ge 2x≥2), find g−1(x)g^{-1}(x)g−1(x).
From y=x−2y = \sqrt{x - 2}y=x−2, swap to x=y−2x = \sqrt{y - 2}x=y−2 and square both sides.
x2=y−2 ⇒ y=x2+2x^2 = y - 2 \;\Rightarrow\; y = x^2 + 2x2=y−2⇒y=x2+2
The input xxx was a square-root output, so x≥0x \ge 0x≥0; thus g−1(x)=x2+2g^{-1}(x) = x^2 + 2g−1(x)=x2+2 for x≥0x \ge 0x≥0.
For a one-to-one function fff, what is f(f−1(8))f(f^{-1}(8))f(f−1(8))?
A function and its inverse undo each other in both directions.
f(f−1(8))=8f(f^{-1}(8)) = 8f(f−1(8))=8
Applying f−1f^{-1}f−1 and then fff returns the original input 888.
If (p,q)(p, q)(p,q) is on the graph of fff with q≠pq \ne pq=p, then (q,p)(q, p)(q,p) is on the graph of f−1f^{-1}f−1. Reflecting (p,q)(p, q)(p,q) perpendicularly across the line y=xy = xy=x and continuing an equal distance beyond it reaches
The line y=xy = xy=x is the perpendicular bisector of the segment from (p,q)(p, q)(p,q) to its swap.
(p,q)→(q,p)(p, q) \to (q, p)(p,q)→(q,p)
Reflecting across the mirror lands exactly on (q,p)(q, p)(q,p), the point on f−1f^{-1}f−1.
A one-to-one function has f(0)=1f(0) = 1f(0)=1, f(1)=3f(1) = 3f(1)=3, f(2)=9f(2) = 9f(2)=9, and f(3)=27f(3) = 27f(3)=27. What is f−1(9)+f−1(3)f^{-1}(9) + f^{-1}(3)f−1(9)+f−1(3)?
Read each inverse value from the table: (2,9)(2, 9)(2,9) gives f−1(9)=2f^{-1}(9) = 2f−1(9)=2, and (1,3)(1, 3)(1,3) gives f−1(3)=1f^{-1}(3) = 1f−1(3)=1.
f−1(9)+f−1(3)=2+1=3f^{-1}(9) + f^{-1}(3) = 2 + 1 = 3f−1(9)+f−1(3)=2+1=3
Each inverse value is the input that produced the given output.
For f(x)=x2+1f(x) = x^2 + 1f(x)=x2+1 with x≥0x \ge 0x≥0, find f−1(x)f^{-1}(x)f−1(x).
From y=x2+1y = x^2 + 1y=x2+1, swap to x=y2+1x = y^2 + 1x=y2+1 and solve for yyy, keeping y≥0y \ge 0y≥0.
y2=x−1 ⇒ y=x−1y^2 = x - 1 \;\Rightarrow\; y = \sqrt{x - 1}y2=x−1⇒y=x−1
So f−1(x)=x−1f^{-1}(x) = \sqrt{x - 1}f−1(x)=x−1.
f(x)=x3+1f(x) = x^3 + 1f(x)=x3+1. Reflecting its graph across y=xy = xy=x gives the graph of f−1(x)=f^{-1}(x) =f−1(x)=
The cube is one-to-one, so the whole reflection is a function. From y=x3+1y = x^3 + 1y=x3+1, swap to x=y3+1x = y^3 + 1x=y3+1.
y3=x−1 ⇒ y=x−13y^3 = x - 1 \;\Rightarrow\; y = \sqrt[3]{x - 1}y3=x−1⇒y=3x−1
So f−1(x)=x−13f^{-1}(x) = \sqrt[3]{x - 1}f−1(x)=3x−1.
When the graph of fff is reflected across y=xy = xy=x to form f−1f^{-1}f−1, an xxx-intercept of fff becomes what feature of f−1f^{-1}f−1?
An xxx-intercept of fff has the form (a,0)(a, 0)(a,0); reflect it across y=xy = xy=x.
(a,0)→(0,a)(a, 0) \to (0, a)(a,0)→(0,a)
A point on the yyy-axis is a yyy-intercept, so the xxx-intercept becomes a yyy-intercept.
fff and ggg are inverses of each other. The graph of ggg passes through (a,b)(a, b)(a,b). The graph of fff must pass through
The graphs of a function and its inverse are reflections across y=xy = xy=x, so a point on one has its swap on the other.
(a,b)→(b,a)(a, b) \to (b, a)(a,b)→(b,a)
So the graph of fff passes through (b,a)(b, a)(b,a).
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