12 multiple-choice questions, progressively harder.
Let f(x)=3x−5f(x) = 3x - 5f(x)=3x−5 and g(x)=x+53g(x) = \dfrac{x + 5}{3}g(x)=3x+5. What is g(f(x))g(f(x))g(f(x))?
Solution
Correct answer: D
Drop the whole rule for fff into ggg and simplify.
g(f(x))=(3x−5)+53=3x3=xg(f(x)) = \frac{(3x - 5) + 5}{3} = \frac{3x}{3} = xg(f(x))=3(3x−5)+5=33x=x
Since it collapses to xxx, ggg undoes fff in this direction.
Which graph passes the horizontal line test?
Correct answer: C
A graph passes the horizontal line test when no horizontal line meets it more than once, which makes the function one-to-one. A slanted line reaches each height exactly once.
f(a)=f(b) ⟹ a=bf(a) = f(b) \;\Longrightarrow\; a = bf(a)=f(b)⟹a=b
A horizontal line sends every input to one output, and both parabolas are met twice by some horizontal line, so those fail.
For f(x)=x2f(x) = x^2f(x)=x2 restricted to x≥0x \ge 0x≥0, find f−1(x)f^{-1}(x)f−1(x).
Correct answer: B
Swap xxx and yyy, then solve, keeping the nonnegative root because the original inputs were ≥0\ge 0≥0.
y=x2 (x≥0) ⟹ x=y2 ⟹ y=xy = x^2 \;(x \ge 0) \;\Longrightarrow\; x = y^2 \;\Longrightarrow\; y = \sqrt{x}y=x2(x≥0)⟹x=y2⟹y=x
The inverse of f(x)=x−2f(x) = \sqrt{x - 2}f(x)=x−2 is f−1(x)=x2+2f^{-1}(x) = x^2 + 2f−1(x)=x2+2 with which restriction?
The range of f(x)=x−2f(x) = \sqrt{x - 2}f(x)=x−2 is y≥0y \ge 0y≥0, and that range becomes the domain of the inverse.
f−1(x)=x2+2,x≥0f^{-1}(x) = x^2 + 2, \quad x \ge 0f−1(x)=x2+2,x≥0
A horizontal line crosses the graph of fff at two points. What does this show?
Correct answer: A
Two points on one horizontal line share the same output but come from different inputs.
f(a)=f(b), a≠bf(a) = f(b), \; a \ne bf(a)=f(b),a=b
That is exactly the failure of the one-to-one condition, so fff has no inverse function.
Which condition confirms that f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 and g(x)=x−12g(x) = \dfrac{x - 1}{2}g(x)=2x−1 are inverses?
Inverses must undo each other in both directions, so both compositions must return xxx.
f(g(x))=2⋅x−12+1=x,g(f(x))=(2x+1)−12=xf(g(x)) = 2 \cdot \frac{x - 1}{2} + 1 = x, \qquad g(f(x)) = \frac{(2x + 1) - 1}{2} = xf(g(x))=2⋅2x−1+1=x,g(f(x))=2(2x+1)−1=x
Because g(f(−4))=4≠−4g(f(-4)) = 4 \ne -4g(f(−4))=4=−4, the functions f(x)=x2f(x) = x^2f(x)=x2 and g(x)=xg(x) = \sqrt{x}g(x)=x are:
The direction g(f(x))=xg(f(x)) = xg(f(x))=x fails for negative inputs.
g(f(x))=x2≠x for x<0g(f(x)) = \sqrt{x^2} \ne x \;\text{ for } x < 0g(f(x))=x2=x for x<0
Since one direction breaks, they are not inverses unless the domain of x2x^2x2 is restricted to x≥0x \ge 0x≥0.
Which restricted domain makes f(x)=(x−3)2f(x) = (x - 3)^2f(x)=(x−3)2 one-to-one and invertible?
The graph turns around at its vertex x=3x = 3x=3, so keep one side of it.
f(x)=(x−3)2, x≥3f(x) = (x - 3)^2, \; x \ge 3f(x)=(x−3)2,x≥3
The sets x≥0x \ge 0x≥0 and x≤6x \le 6x≤6 each contain inputs on both sides of 333 (for example 222 and 444 give the same output), so they are not one-to-one.
Let f(x)=x3+2f(x) = \dfrac{x}{3} + 2f(x)=3x+2 and g(x)=3x−6g(x) = 3x - 6g(x)=3x−6. Compute f(g(x))f(g(x))f(g(x)).
Substitute ggg into fff and simplify.
f(g(x))=3x−63+2=(x−2)+2=xf(g(x)) = \frac{3x - 6}{3} + 2 = (x - 2) + 2 = xf(g(x))=33x−6+2=(x−2)+2=x
The horizontal line test tells you whether a function is ___; the vertical line test tells you whether a graph is a ___.
The horizontal line test checks one input per output (one-to-one); the vertical line test checks one output per input (a function).
horizontal→one-to-one,vertical→function\text{horizontal} \to \text{one-to-one}, \qquad \text{vertical} \to \text{function}horizontal→one-to-one,vertical→function
The functions fff and ggg are inverses. If f(5)=−2f(5) = -2f(5)=−2, then g(−2)=g(-2) = {}g(−2)=?
Inverses reverse each pair, so (5,−2)(5, -2)(5,−2) becomes (−2,5)(-2, 5)(−2,5).
f(5)=−2 ⟹ g(−2)=5f(5) = -2 \;\Longrightarrow\; g(-2) = 5f(5)=−2⟹g(−2)=5
Which set of pairs represents a one-to-one function?
A one-to-one function needs distinct inputs and distinct outputs.
{(1,4),(2,5),(3,6)}\{(1, 4), (2, 5), (3, 6)\}{(1,4),(2,5),(3,6)}
The first two sets repeat an output, and the last repeats the input 111 (so it is not even a function).
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