12 multiple-choice questions, progressively harder.
Let f(x)=1x−3f(x) = \dfrac{1}{x - 3}f(x)=x−31 and g(x)=1x+3g(x) = \dfrac{1}{x} + 3g(x)=x1+3. Compute f(g(x))f(g(x))f(g(x)).
Solution
Correct answer: D
Substitute ggg into fff; the added 333 and the subtracted 333 cancel inside the denominator.
f(g(x))=1(1x+3)−3=11x=xf(g(x)) = \frac{1}{\left(\frac{1}{x} + 3\right) - 3} = \frac{1}{\frac{1}{x}} = xf(g(x))=(x1+3)−31=x11=x
So ggg undoes fff in this direction.
Using V(t)=50+4tV(t) = 50 + 4tV(t)=50+4t from the tank problem, how many minutes give a volume of 868686 litres?
Correct answer: A
Apply the inverse t=V−504t = \dfrac{V - 50}{4}t=4V−50 at V=86V = 86V=86, or just solve directly.
86=50+4t ⟹ 4t=36 ⟹ t=986 = 50 + 4t \;\Longrightarrow\; 4t = 36 \;\Longrightarrow\; t = 986=50+4t⟹4t=36⟹t=9
Let f(x)=x2f(x) = x^2f(x)=x2 on all real numbers and g(x)=xg(x) = \sqrt{x}g(x)=x. Which statement is correct?
Squaring destroys the sign of a negative input, so one direction fails.
f(g(x))=(x)2=x,g(f(−3))=9=3≠−3f(g(x)) = \left(\sqrt{x}\right)^2 = x, \qquad g(f(-3)) = \sqrt{9} = 3 \ne -3f(g(x))=(x)2=x,g(f(−3))=9=3=−3
Because g(f(x))=xg(f(x)) = xg(f(x))=x breaks for negatives, the two are not full inverses.
Find the inverse of f(x)=2x+1xf(x) = \dfrac{2x + 1}{x}f(x)=x2x+1.
Correct answer: C
First split the rule: 2x+1x=2+1x\dfrac{2x + 1}{x} = 2 + \dfrac{1}{x}x2x+1=2+x1. Swap xxx and yyy, then solve.
x=2+1y ⟹ 1y=x−2 ⟹ y=1x−2x = 2 + \frac{1}{y} \;\Longrightarrow\; \frac{1}{y} = x - 2 \;\Longrightarrow\; y = \frac{1}{x - 2}x=2+y1⟹y1=x−2⟹y=x−21
The function f(x)=1x+5f(x) = \dfrac{1}{x} + 5f(x)=x1+5 has range all real numbers except 555. What is the domain of f−1f^{-1}f−1?
Correct answer: B
The domain of the inverse is the range of the original function.
domain(f−1)=range(f)=all real numbers except 5\text{domain}(f^{-1}) = \text{range}(f) = \text{all real numbers except } 5domain(f−1)=range(f)=all real numbers except 5
After checking both compositions, the functions f(x)=5x+3f(x) = 5x + 3f(x)=5x+3 and g(x)=x−35g(x) = \dfrac{x - 3}{5}g(x)=5x−3 are:
Test both directions.
g(f(x))=(5x+3)−35=x,f(g(x))=5⋅x−35+3=xg(f(x)) = \frac{(5x + 3) - 3}{5} = x, \qquad f(g(x)) = 5 \cdot \frac{x - 3}{5} + 3 = xg(f(x))=5(5x+3)−3=x,f(g(x))=5⋅5x−3+3=x
Both return xxx, so they are inverses.
A function's graph is a parabola opening upward. How many times can a horizontal line drawn above the vertex cross it?
Above the vertex the horizontal line meets both branches of the parabola.
x2=c ⟹ x=±cx^2 = c \;\Longrightarrow\; x = \pm\sqrt{c}x2=c⟹x=±c
Two crossings mean two inputs share an output, so the function is not one-to-one.
Which pair of functions are inverses of each other?
Test the promising pair in both directions.
g(f(x))=(x−8)+8=x,f(g(x))=(x+8)−8=xg(f(x)) = (x - 8) + 8 = x, \qquad f(g(x)) = (x + 8) - 8 = xg(f(x))=(x−8)+8=x,f(g(x))=(x+8)−8=x
The doubling pair gives g(f(x))=4xg(f(x)) = 4xg(f(x))=4x, the 8−x8 - x8−x pair gives 16−x16 - x16−x, and x2x^2x2 with x\sqrt{x}x fails for negatives, so only x−8x - 8x−8 and x+8x + 8x+8 work.
If fff is one-to-one and f(a)=f(b)f(a) = f(b)f(a)=f(b), what must be true?
One-to-one means different inputs always give different outputs, so equal outputs can only come from equal inputs.
f(a)=f(b) ⟹ a=bf(a) = f(b) \;\Longrightarrow\; a = bf(a)=f(b)⟹a=b
Let f(x)=4x−1f(x) = 4x - 1f(x)=4x−1 with inverse g(x)=x+14g(x) = \dfrac{x + 1}{4}g(x)=4x+1. Compute f(g(x))f(g(x))f(g(x)).
Substitute ggg into fff and simplify.
f(g(x))=4⋅x+14−1=(x+1)−1=xf(g(x)) = 4 \cdot \frac{x + 1}{4} - 1 = (x + 1) - 1 = xf(g(x))=4⋅4x+1−1=(x+1)−1=x
Why must f(x)=x2f(x) = x^2f(x)=x2 be restricted before it has an inverse function?
On all real numbers two inputs share each positive output.
f(3)=9=f(−3)f(3) = 9 = f(-3)f(3)=9=f(−3)
Until the domain is restricted (for example to x≥0x \ge 0x≥0), one output would have to map back to two inputs, which no function can do.
A game converts dollars to coins with C(d)=15dC(d) = 15dC(d)=15d. Which function converts coins CCC back to dollars?
Undo "multiply by 15" by dividing by 151515.
C=15d ⟹ d=C15C = 15d \;\Longrightarrow\; d = \frac{C}{15}C=15d⟹d=15C
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