12 multiple-choice questions, progressively harder.
For f(x)=1xf(x) = \dfrac{1}{x}f(x)=x1, 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: A
Combine the fractions over x(x+h)x(x + h)x(x+h), then divide by hhh.
1x+h−1x=−hx(x+h),1h⋅−hx(x+h)=−1x(x+h)\frac{1}{x + h} - \frac{1}{x} = \frac{-h}{x(x + h)}, \qquad \frac{1}{h} \cdot \frac{-h}{x(x + h)} = \frac{-1}{x(x + h)}x+h1−x1=x(x+h)−h,h1⋅x(x+h)−h=x(x+h)−1
The hhh cancels for h≠0h \ne 0h=0.
If f(x)=2x+1f(x) = 2x + 1f(x)=2x+1, which expression equals f(f(x))f(f(x))f(f(x))?
Correct answer: B
Substitute f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 into the slot of fff.
f(f(x))=2(2x+1)+1=4x+2+1=4x+3f(f(x)) = 2(2x + 1) + 1 = 4x + 2 + 1 = 4x + 3f(f(x))=2(2x+1)+1=4x+2+1=4x+3
Multiply through, then add.
Are f(x)=x2−1x−1f(x) = \dfrac{x^2 - 1}{x - 1}f(x)=x−1x2−1 and g(x)=x+1g(x) = x + 1g(x)=x+1 the same function?
Factor and cancel for x≠1x \ne 1x=1: (x−1)(x+1)x−1=x+1\dfrac{(x - 1)(x + 1)}{x - 1} = x + 1x−1(x−1)(x+1)=x+1.
f(1)=00 undefined,g(1)=2f(1) = \tfrac{0}{0} \ \text{undefined}, \qquad g(1) = 2f(1)=00 undefined,g(1)=2
Different domains means different functions.
For f(x)=x2f(x) = x^2f(x)=x2, simplify f(a)−f(b)a−b\dfrac{f(a) - f(b)}{a - b}a−bf(a)−f(b) for a≠ba \ne ba=b.
Factor the difference of squares in the numerator.
a2−b2a−b=(a−b)(a+b)a−b=a+b\frac{a^2 - b^2}{a - b} = \frac{(a - b)(a + b)}{a - b} = a + ba−ba2−b2=a−b(a−b)(a+b)=a+b
The cancellation needs a≠ba \ne ba=b.
Let f(x)=x+1f(x) = x + 1f(x)=x+1 for x<0x < 0x<0, f(x)=x2f(x) = x^2f(x)=x2 for 0≤x≤20 \le x \le 20≤x≤2, and f(x)=5f(x) = 5f(x)=5 for x>2x > 2x>2. What is f(−2)f(-2)f(−2)?
Correct answer: D
Since −2<0-2 < 0−2<0, use the first rule.
f(−2)=−2+1=−1f(-2) = -2 + 1 = -1f(−2)=−2+1=−1
Read the condition to pick the case that owns the input.
If f(x)=1x−2f(x) = \dfrac{1}{x - 2}f(x)=x−21, which expression equals f(x+2)f(x + 2)f(x+2)?
Correct answer: C
Put the whole input x+2x + 2x+2 in the slot.
f(x+2)=1(x+2)−2=1xf(x + 2) = \frac{1}{(x + 2) - 2} = \frac{1}{x}f(x+2)=(x+2)−21=x1
The +2+2+2 and −2-2−2 cancel inside the denominator.
For f(x)=x2f(x) = x^2f(x)=x2, simplify f(3+h)−f(3)h\dfrac{f(3 + h) - f(3)}{h}hf(3+h)−f(3) for h≠0h \ne 0h=0.
Expand (3+h)2(3 + h)^2(3+h)2 and subtract 999.
(9+6h+h2)−9h=6h+h2h=6+h\frac{(9 + 6h + h^2) - 9}{h} = \frac{6h + h^2}{h} = 6 + hh(9+6h+h2)−9=h6h+h2=6+h
This is the difference quotient centered at x=3x = 3x=3.
For f(x)=mx+cf(x) = mx + cf(x)=mx+c, when is f(a+b)=f(a)+f(b)f(a + b) = f(a) + f(b)f(a+b)=f(a)+f(b) true for all aaa and bbb?
Compute both sides.
f(a+b)=ma+mb+c,f(a)+f(b)=ma+mb+2cf(a + b) = ma + mb + c, \qquad f(a) + f(b) = ma + mb + 2cf(a+b)=ma+mb+c,f(a)+f(b)=ma+mb+2c
These are equal exactly when c=2cc = 2cc=2c, that is c=0c = 0c=0, the lines through the origin.
Are f(x)=∣x∣f(x) = |x|f(x)=∣x∣ and g(x)=x2g(x) = \sqrt{x^2}g(x)=x2 the same function?
The principal square root of x2x^2x2 is the nonnegative number whose square is x2x^2x2.
x2=∣x∣ for every real x\sqrt{x^2} = |x| \ \text{for every real } xx2=∣x∣ for every real x
Both have domain all reals and equal values, so they are the same function.
Let f(x)=x2f(x) = x^2f(x)=x2 with the declared domain x≥0x \ge 0x≥0. Is f(−3)f(-3)f(−3) defined?
The declared domain is part of the function, and −3<0-3 < 0−3<0.
−3∉{x:x≥0}-3 \notin \{x : x \ge 0\}−3∈/{x:x≥0}
So f(−3)f(-3)f(−3) is undefined even though the formula x2x^2x2 would give 999.
For f(x)=3xf(x) = 3xf(x)=3x and any constant kkk, which expression equals f(kx)f(kx)f(kx)?
Substitute kxkxkx into the rule.
f(kx)=3(kx)=k(3x)=k f(x)f(kx) = 3(kx) = k(3x) = k\,f(x)f(kx)=3(kx)=k(3x)=kf(x)
Scaling the input by kkk scales the output by kkk for f(x)=mxf(x) = mxf(x)=mx.
For f(x)=x2+2xf(x) = x^2 + 2xf(x)=x2+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.
Expand f(x+h)=(x+h)2+2(x+h)f(x + h) = (x + h)^2 + 2(x + h)f(x+h)=(x+h)2+2(x+h), then subtract f(x)=x2+2xf(x) = x^2 + 2xf(x)=x2+2x.
f(x+h)−f(x)=(x2+2xh+h2+2x+2h)−(x2+2x)=2xh+h2+2hf(x + h) - f(x) = (x^2 + 2xh + h^2 + 2x + 2h) - (x^2 + 2x) = 2xh + h^2 + 2hf(x+h)−f(x)=(x2+2xh+h2+2x+2h)−(x2+2x)=2xh+h2+2h
Now divide by hhh.
2xh+h2+2hh=2x+h+2\frac{2xh + h^2 + 2h}{h} = 2x + h + 2h2xh+h2+2h=2x+h+2
The 2x2x2x term contributes the +2+2+2.
Reset this practice set?
This clears every answer you have given and starts the set again from question 1.