12 multiple-choice questions, progressively harder.
For f(x)=3x−6f(x) = 3x - 6f(x)=3x−6, the point (4,6)(4, 6)(4,6) lies on the graph. Which point lies on the graph of f−1f^{-1}f−1?
Solution
Correct answer: B
First check the point: f(4)=3(4)−6=6f(4) = 3(4) - 6 = 6f(4)=3(4)−6=6, so (4,6)(4, 6)(4,6) is on fff. Reflect across y=xy = xy=x by swapping coordinates.
(4,6)→(6,4)(4, 6) \to (6, 4)(4,6)→(6,4)
So (6,4)(6, 4)(6,4) is on the graph of f−1f^{-1}f−1.
A line has slope 14\dfrac{1}{4}41. Its inverse is also a line. What is the slope of the inverse?
Correct answer: A
Reflecting across y=xy = xy=x swaps the run and the rise, so the slope becomes its reciprocal.
slope of inverse=11/4=4\text{slope of inverse} = \frac{1}{1/4} = 4slope of inverse=1/41=4
A gentle line of slope 14\tfrac1441 reflects to a steep line of slope 444.
The function f(x)=xf(x) = \sqrt{x}f(x)=x has domain x≥0x \ge 0x≥0 and range y≥0y \ge 0y≥0. Its inverse is f−1(x)=x2f^{-1}(x) = x^2f−1(x)=x2 with which restriction?
Correct answer: C
The domain of the inverse is the range of fff, which is y≥0y \ge 0y≥0, so f−1f^{-1}f−1 accepts inputs x≥0x \ge 0x≥0.
domain of f−1=range of f: x≥0\text{domain of } f^{-1} = \text{range of } f: \; x \ge 0domain of f−1=range of f:x≥0
Without the restriction, x2x^2x2 would not be one-to-one and could not be the inverse.
For f(x)=x3f(x) = x^3f(x)=x3, is a domain restriction needed before reflecting across y=xy = xy=x to get the graph of a function?
As xxx increases, x3x^3x3 always increases, so every horizontal line meets the graph exactly once.
a≠b ⇒ a3≠b3a \ne b \;\Rightarrow\; a^3 \ne b^3a=b⇒a3=b3
The cube is one-to-one, so its full reflection is already a function (the cube-root curve); no restriction is needed.
A function fff has domain −2≤x≤6-2 \le x \le 6−2≤x≤6 and range 1≤y≤91 \le y \le 91≤y≤9. What is the range of f−1f^{-1}f−1?
Correct answer: D
Reflecting across y=xy = xy=x trades the axes, so the range of the inverse is the domain of fff.
range of f−1=domain of f=−2≤y≤6\text{range of } f^{-1} = \text{domain of } f = -2 \le y \le 6range of f−1=domain of f=−2≤y≤6
The old inputs become the new outputs.
The line f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 passes through (0,1)(0, 1)(0,1) and (1,3)(1, 3)(1,3). Which pair of points lies on the graph of f−1f^{-1}f−1?
Swap the coordinates of each point on fff.
(0,1)→(1,0),(1,3)→(3,1)(0, 1) \to (1, 0), \qquad (1, 3) \to (3, 1)(0,1)→(1,0),(1,3)→(3,1)
So f−1f^{-1}f−1 passes through (1,0)(1, 0)(1,0) and (3,1)(3, 1)(3,1).
The graph of f−1f^{-1}f−1 passes through the point (7,3)(7, 3)(7,3). Which point lies on the graph of fff?
Reflect the point on f−1f^{-1}f−1 back across y=xy = xy=x to land on fff.
(7,3)→(3,7)(7, 3) \to (3, 7)(7,3)→(3,7)
So (3,7)(3, 7)(3,7) is on the graph of fff.
Restricting f(x)=x2f(x) = x^2f(x)=x2 to x≥0x \ge 0x≥0 keeps which part of the parabola?
The inputs x≥0x \ge 0x≥0 are the nonnegative ones, so only the branch to the right of the vertex survives.
x≥0 ⇒ right branch, in the first quadrantx \ge 0 \;\Rightarrow\; \text{right branch, in the first quadrant}x≥0⇒right branch, in the first quadrant
That branch passes the horizontal line test, so it can be reflected into a function.
After restricting to x≥0x \ge 0x≥0 and reflecting across y=xy = xy=x, the graph of f(x)=x2f(x) = x^2f(x)=x2 becomes the graph of
The restricted branch is one-to-one, and reflecting it across y=xy = xy=x inverts the rule.
y=x2, x≥0 ⟶ y=xy = x^2, \; x \ge 0 \;\longrightarrow\; y = \sqrt{x}y=x2,x≥0⟶y=x
The square-root curve is the reflection of the right branch.
The horizontal line y=cy = cy=c is reflected across y=xy = xy=x. What does it become?
Every point (t,c)(t, c)(t,c) on the horizontal line swaps to (c,t)(c, t)(c,t).
(t,c)→(c,t)(t, c) \to (c, t)(t,c)→(c,t)
The collection of all (c,t)(c, t)(c,t) is the vertical line x=cx = cx=c, so horizontal lines and vertical lines are mirror images.
If f(5)=5f(5) = 5f(5)=5, what is f−1(5)f^{-1}(5)f−1(5)?
The point (5,5)(5, 5)(5,5) is on fff and lies on the line y=xy = xy=x, so it is a fixed point of the reflection.
(5,5)→(5,5) ⇒ f−1(5)=5(5, 5) \to (5, 5) \;\Rightarrow\; f^{-1}(5) = 5(5,5)→(5,5)⇒f−1(5)=5
A value the function leaves unchanged is left unchanged by the inverse.
The graph of fff lies entirely in the first quadrant (x≥0x \ge 0x≥0 and y≥0y \ge 0y≥0). Its reflection across y=xy = xy=x lies in
Swapping two nonnegative coordinates gives two nonnegative coordinates.
(a,b)→(b,a),a,b≥0(a, b) \to (b, a), \quad a, b \ge 0(a,b)→(b,a),a,b≥0
Both points are in the first quadrant, so the reflection stays there.
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