12 multiple-choice questions, progressively harder.
f(x)=4x−1f(x) = 4x - 1f(x)=4x−1 passes through (2,7)(2, 7)(2,7). Which point lies on f−1f^{-1}f−1, and what is f−1(7)f^{-1}(7)f−1(7)?
Solution
Correct answer: A
Check the point: f(2)=4(2)−1=7f(2) = 4(2) - 1 = 7f(2)=4(2)−1=7. Reflect (2,7)(2, 7)(2,7) across y=xy = xy=x by swapping coordinates.
(2,7)→(7,2) ⇒ f−1(7)=2(2, 7) \to (7, 2) \;\Rightarrow\; f^{-1}(7) = 2(2,7)→(7,2)⇒f−1(7)=2
The swapped point (7,2)(7, 2)(7,2) gives the inverse value directly.
f(x)=x2+3f(x) = \dfrac{x}{2} + 3f(x)=2x+3. Its inverse is a line. What is the slope of the inverse?
Correct answer: C
The slope of fff is 12\tfrac{1}{2}21, and reflecting across y=xy = xy=x turns a slope into its reciprocal.
slope of f−1=11/2=2\text{slope of } f^{-1} = \frac{1}{1/2} = 2slope of f−1=1/21=2
The gentle line reflects to a steeper line of slope 222.
h(x)=x2−4h(x) = x^2 - 4h(x)=x2−4 for x≥0x \ge 0x≥0. What is the domain of h−1h^{-1}h−1?
Correct answer: B
The domain of the inverse is the range of hhh. On x≥0x \ge 0x≥0 the smallest output is h(0)=−4h(0) = -4h(0)=−4, and it increases from there.
range of h: y≥−4 ⇒ domain of h−1: x≥−4\text{range of } h: \; y \ge -4 \;\Rightarrow\; \text{domain of } h^{-1}: \; x \ge -4range of h:y≥−4⇒domain of h−1:x≥−4
The outputs of hhh become the allowed inputs of h−1h^{-1}h−1.
f(x)=x3−8f(x) = x^3 - 8f(x)=x3−8. Reflecting its graph across y=xy = xy=x gives f−1(x)=f^{-1}(x) =f−1(x)=
The cube is one-to-one, so the full reflection is a function. From y=x3−8y = x^3 - 8y=x3−8, swap to x=y3−8x = y^3 - 8x=y3−8.
y3=x+8 ⇒ y=x+83y^3 = x + 8 \;\Rightarrow\; y = \sqrt[3]{x + 8}y3=x+8⇒y=3x+8
So f−1(x)=x+83f^{-1}(x) = \sqrt[3]{x + 8}f−1(x)=3x+8.
f(x)=x+1f(x) = \sqrt{x} + 1f(x)=x+1. Reflecting its graph across y=xy = xy=x gives f−1(x)=f^{-1}(x) =f−1(x)=
Correct answer: D
From y=x+1y = \sqrt{x} + 1y=x+1, swap to x=y+1x = \sqrt{y} + 1x=y+1, so y=x−1\sqrt{y} = x - 1y=x−1, then square.
y=(x−1)2y = (x - 1)^2y=(x−1)2
Since y=x−1≥0\sqrt{y} = x - 1 \ge 0y=x−1≥0, the domain is x≥1x \ge 1x≥1, so f−1(x)=(x−1)2f^{-1}(x) = (x - 1)^2f−1(x)=(x−1)2 for x≥1x \ge 1x≥1.
A line makes a 30∘30^\circ30∘ angle with the positive xxx-axis. Its inverse's graph makes what angle with the positive xxx-axis?
Reflecting across the 45∘45^\circ45∘ line y=xy = xy=x sends an angle θ\thetaθ to 90∘−θ90^\circ - \theta90∘−θ.
90∘−30∘=60∘90^\circ - 30^\circ = 60^\circ90∘−30∘=60∘
So the inverse's line makes a 60∘60^\circ60∘ angle.
fff is a one-to-one function. Which statement is always true?
Running fff and then f−1f^{-1}f−1 returns the starting input.
f−1(f(a))=af^{-1}(f(a)) = af−1(f(a))=a
The superscript −1-1−1 names the inverse, not the reciprocal, so f−1(x)≠1f(x)f^{-1}(x) \ne \tfrac{1}{f(x)}f−1(x)=f(x)1 in general, and fff need not equal its own inverse.
The point (3,−2)(3, -2)(3,−2) is on the graph of fff. Its reflection across y=xy = xy=x lands in which quadrant?
Swap the coordinates to reflect across y=xy = xy=x.
(3,−2)→(−2,3)(3, -2) \to (-2, 3)(3,−2)→(−2,3)
The point (−2,3)(-2, 3)(−2,3) has a negative xxx and a positive yyy, so it is in quadrant II.
A function fff has an xxx-intercept at (3,0)(3, 0)(3,0) and a yyy-intercept at (0,−6)(0, -6)(0,−6). After reflecting across y=xy = xy=x, the intercepts of f−1f^{-1}f−1 are
Swap the coordinates of each intercept; an xxx-intercept and a yyy-intercept trade roles.
(3,0)→(0,3),(0,−6)→(−6,0)(3, 0) \to (0, 3), \qquad (0, -6) \to (-6, 0)(3,0)→(0,3),(0,−6)→(−6,0)
So f−1f^{-1}f−1 has xxx-intercept (−6,0)(-6, 0)(−6,0) and yyy-intercept (0,3)(0, 3)(0,3).
Reflect the point (−5,−1)(-5, -1)(−5,−1) across the line y=xy = xy=x.
Swap the two coordinates, keeping each sign attached to its number.
(−5,−1)→(−1,−5)(-5, -1) \to (-1, -5)(−5,−1)→(−1,−5)
So the reflection is (−1,−5)(-1, -5)(−1,−5).
f(x)=1x−1f(x) = \dfrac{1}{x - 1}f(x)=x−11 passes through (2,1)(2, 1)(2,1). Which point lies on the graph of f−1f^{-1}f−1?
Swap the coordinates of the point on fff.
(2,1)→(1,2)(2, 1) \to (1, 2)(2,1)→(1,2)
Check with f−1(x)=1x+1f^{-1}(x) = \dfrac{1}{x} + 1f−1(x)=x1+1: f−1(1)=1+1=2f^{-1}(1) = 1 + 1 = 2f−1(1)=1+1=2, so (1,2)(1, 2)(1,2) is on the graph of f−1f^{-1}f−1.
The graphs of fff and f−1f^{-1}f−1 are reflections across y=xy = xy=x. Which statement is guaranteed?
The defining property of the reflection is that it swaps coordinates.
(a,b) on f ⟺ (b,a) on f−1(a, b) \text{ on } f \iff (b, a) \text{ on } f^{-1}(a,b) on f⟺(b,a) on f−1
The graphs may intersect (on y=xy = xy=x, for instance), and they are generally not parallel, so only the coordinate-swap statement is guaranteed.
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