12 multiple-choice questions, progressively harder.
For f(x)=x2−1f(x) = x^2 - 1f(x)=x2−1, simplify f(x+h)−f(x)h\dfrac{f(x + h) - f(x)}{h}hf(x+h)−f(x) for h≠0h \ne 0h=0.
Solution
Correct answer: B
The constant −1-1−1 cancels between f(x+h)f(x + h)f(x+h) and f(x)f(x)f(x).
[(x+h)2−1]−[x2−1]h=2xh+h2h=2x+h\frac{[(x + h)^2 - 1] - [x^2 - 1]}{h} = \frac{2xh + h^2}{h} = 2x + hh[(x+h)2−1]−[x2−1]=h2xh+h2=2x+h
Only the squared term survives.
For f(x)=5−2xf(x) = 5 - 2xf(x)=5−2x, simplify f(x+h)−f(x)h\dfrac{f(x + h) - f(x)}{h}hf(x+h)−f(x) for h≠0h \ne 0h=0.
Correct answer: D
Subtract the values, then divide.
[5−2(x+h)]−[5−2x]h=−2hh=−2\frac{[5 - 2(x + h)] - [5 - 2x]}{h} = \frac{-2h}{h} = -2h[5−2(x+h)]−[5−2x]=h−2h=−2
The difference quotient of a line is its slope, here −2-2−2.
If f(x)=x2f(x) = x^2f(x)=x2, which expression equals f(x−h)f(x - h)f(x−h) expanded?
Correct answer: A
Square the whole input x−hx - hx−h.
f(x−h)=(x−h)2=x2−2xh+h2f(x - h) = (x - h)^2 = x^2 - 2xh + h^2f(x−h)=(x−h)2=x2−2xh+h2
The middle term is negative because the input has a minus sign.
If f(x)=x2+xf(x) = x^2 + xf(x)=x2+x, which expression equals f(−x)f(-x)f(−x)?
Correct answer: C
Substitute −x-x−x, wrapping it in parentheses.
f(−x)=(−x)2+(−x)=x2−xf(-x) = (-x)^2 + (-x) = x^2 - xf(−x)=(−x)2+(−x)=x2−x
The squared term stays positive; the linear term flips sign.
What is the natural domain of f(x)=x2xf(x) = \dfrac{x^2}{x}f(x)=xx2?
The denominator xxx cannot be zero.
x≠0x \ne 0x=0
It simplifies to xxx for x≠0x \ne 0x=0, but the value 000 is still excluded from the domain.
Let f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 for x≤1x \le 1x≤1 and f(x)=4f(x) = 4f(x)=4 for x>1x > 1x>1. What is f(1)f(1)f(1)?
Since 1≤11 \le 11≤1, use the first rule.
f(1)=2(1)+1=3f(1) = 2(1) + 1 = 3f(1)=2(1)+1=3
The constant 444 needs x>1x > 1x>1, which 111 does not satisfy.
Let f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 for x≤1x \le 1x≤1 and f(x)=4f(x) = 4f(x)=4 for x>1x > 1x>1. What is f(3)f(3)f(3)?
Since 3>13 > 13>1, use the constant rule.
f(3)=4f(3) = 4f(3)=4
The first rule only covers x≤1x \le 1x≤1.
For f(x)=x2f(x) = x^2f(x)=x2, simplify f(2+h)−f(2)h\dfrac{f(2 + h) - f(2)}{h}hf(2+h)−f(2) for h≠0h \ne 0h=0.
Expand (2+h)2(2 + h)^2(2+h)2 and subtract 444.
(4+4h+h2)−4h=4h+h2h=4+h\frac{(4 + 4h + h^2) - 4}{h} = \frac{4h + h^2}{h} = 4 + hh(4+4h+h2)−4=h4h+h2=4+h
This is the difference quotient centered at x=2x = 2x=2.
If f(x)=3x−2f(x) = 3x - 2f(x)=3x−2, for which input is f(x)=xf(x) = xf(x)=x?
Set the output equal to the input and solve.
3x−2=x ⇒ 2x=2 ⇒ x=13x - 2 = x \ \Rightarrow \ 2x = 2 \ \Rightarrow \ x = 13x−2=x ⇒ 2x=2 ⇒ x=1
An input equal to its own output is a fixed point.
If f(x)=1x+1f(x) = \dfrac{1}{x + 1}f(x)=x+11, what is f(2)f(2)f(2)?
Put 222 into the slot; the whole x+1x + 1x+1 is the denominator.
f(2)=12+1=13f(2) = \frac{1}{2 + 1} = \frac{1}{3}f(2)=2+11=31
Add inside the denominator first.
If f(x)=x2+1f(x) = x^2 + 1f(x)=x2+1, what is f(2)f(\sqrt{2})f(2)?
Square the input 2\sqrt{2}2, which gives 222.
f(2)=(2)2+1=2+1=3f(\sqrt{2}) = (\sqrt{2})^2 + 1 = 2 + 1 = 3f(2)=(2)2+1=2+1=3
Squaring undoes the square root here.
Which pair of rules defines the SAME function?
For all real xxx, x2=∣x∣\sqrt{x^2} = |x|x2=∣x∣, and both sides are defined everywhere.
x2=∣x∣ for every real x\sqrt{x^2} = |x| \ \text{for every real } xx2=∣x∣ for every real x
Each other pair agrees only on inputs allowed by both domains, but their domains differ (excluding x=0x = 0x=0, all x<0x < 0x<0, or x=1x = 1x=1), so those are different functions.
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