12 multiple-choice questions, progressively harder.
Find the inverse of f(x)=2x+13f(x) = \dfrac{2x + 1}{3}f(x)=32x+1.
Solution
Correct answer: B
Solve y=2x+13y = \frac{2x + 1}{3}y=32x+1 for xxx.
3y=2x+1 ⇒ x=3y−123y = 2x + 1 \;\Rightarrow\; x = \frac{3y - 1}{2}3y=2x+1⇒x=23y−1
So f−1(x)=3x−12f^{-1}(x) = \frac{3x - 1}{2}f−1(x)=23x−1.
For f(x)=x2f(x) = x^2f(x)=x2 on x≥0x \ge 0x≥0, a student writes f−1(x)=xf^{-1}(x) = \sqrt{x}f−1(x)=x and checks only f(f−1(x))=(x)2=xf(f^{-1}(x)) = (\sqrt{x})^2 = xf(f−1(x))=(x)2=x. Is the check complete?
An inverse needs both compositions, so the second one still must be checked.
f−1(f(x))=x2=∣x∣=x because x≥0f^{-1}(f(x)) = \sqrt{x^2} = |x| = x \text{ because } x \ge 0f−1(f(x))=x2=∣x∣=x because x≥0
Here it does hold, precisely because the domain was restricted to x≥0x \ge 0x≥0; on all reals this side would fail, which is why one identity is never enough on its own.
If f−1(x)=2x+3f^{-1}(x) = 2x + 3f−1(x)=2x+3, then f(x)=f(x) =f(x)=?
Correct answer: A
fff is the inverse of f−1f^{-1}f−1, so undo 2x+32x + 32x+3.
y=2x+3 ⇒ x=y−32y = 2x + 3 \;\Rightarrow\; x = \frac{y - 3}{2}y=2x+3⇒x=2y−3
So f(x)=x−32f(x) = \frac{x - 3}{2}f(x)=2x−3.
f(x)=xx−2f(x) = \dfrac{x}{x - 2}f(x)=x−2x on x≠2x \ne 2x=2. Find f−1(x)f^{-1}(x)f−1(x).
Correct answer: C
Solve y=xx−2y = \frac{x}{x - 2}y=x−2x for xxx.
y(x−2)=x ⇒ x(y−1)=2y ⇒ x=2yy−1y(x - 2) = x \;\Rightarrow\; x(y - 1) = 2y \;\Rightarrow\; x = \frac{2y}{y - 1}y(x−2)=x⇒x(y−1)=2y⇒x=y−12y
So f−1(x)=2xx−1f^{-1}(x) = \frac{2x}{x - 1}f−1(x)=x−12x.
The parabola f(x)=(x+1)2f(x) = (x + 1)^2f(x)=(x+1)2 has its vertex at x=−1x = -1x=−1. Which restriction makes fff one-to-one on the largest interval where it is increasing?
Correct answer: D
The parabola turns at x=−1x = -1x=−1 and rises to its right, so it increases on [−1,∞)[-1, \infty)[−1,∞).
f increasing on x≥−1, the largest such intervalf \text{ increasing on } x \ge -1, \text{ the largest such interval}f increasing on x≥−1, the largest such interval
The side x≤−1x \le -1x≤−1 is decreasing, all reals is not one-to-one, and x≥0x \ge 0x≥0 is only part of the increasing branch.
Two functions ggg and hhh are both inverses of fff. In the chain g=g∘(f∘h)=(g∘f)∘h=hg = g \circ (f \circ h) = (g \circ f) \circ h = hg=g∘(f∘h)=(g∘f)∘h=h, which property makes the middle step valid?
Regrouping g∘(f∘h)g \circ (f \circ h)g∘(f∘h) into (g∘f)∘h(g \circ f) \circ h(g∘f)∘h is exactly the associative law.
g∘(f∘h)=(g∘f)∘hg \circ (f \circ h) = (g \circ f) \circ hg∘(f∘h)=(g∘f)∘h
Composition is associative but not commutative, so it is associativity that drives the uniqueness proof.
If f(x)=1x+2f(x) = \dfrac{1}{x + 2}f(x)=x+21 with codomain its range, what is rangef\operatorname{range} frangef, and hence domf−1\operatorname{dom} f^{-1}domf−1?
A fraction with numerator 111 never equals 000, and reaches every other value.
1x+2=0 has no solution ⇒ rangef=R∖{0}\frac{1}{x + 2} = 0 \text{ has no solution} \;\Rightarrow\; \operatorname{range} f = \mathbb{R} \setminus \{0\}x+21=0 has no solution⇒rangef=R∖{0}
So domf−1\operatorname{dom} f^{-1}domf−1 is all reals except 000. The value −2-2−2 is excluded from the domain of fff, not from its range.
Find f−1(x)f^{-1}(x)f−1(x) for f(x)=x+3f(x) = \sqrt{x + 3}f(x)=x+3 with domain x≥−3x \ge -3x≥−3 and range y≥0y \ge 0y≥0.
Solve y=x+3y = \sqrt{x + 3}y=x+3 and take the domain to be the range of fff.
y2=x+3 ⇒ x=y2−3,domf−1=rangef=[0,∞)y^2 = x + 3 \;\Rightarrow\; x = y^2 - 3, \qquad \operatorname{dom} f^{-1} = \operatorname{range} f = [0, \infty)y2=x+3⇒x=y2−3,domf−1=rangef=[0,∞)
So f−1(x)=x2−3f^{-1}(x) = x^2 - 3f−1(x)=x2−3 on x≥0x \ge 0x≥0; allowing all reals would admit inputs the inverse cannot receive.
f(x)=x3−3xf(x) = x^3 - 3xf(x)=x3−3x is not one-to-one on R\mathbb{R}R; for instance f(0)=f(3)=0f(0) = f(\sqrt{3}) = 0f(0)=f(3)=0. Which statement is correct?
The two inputs 000 and 3\sqrt{3}3 share the output 000, so fff is not one-to-one on R\mathbb{R}R.
f(0)=0=f(3) ⇒ no inverse on Rf(0) = 0 = f(\sqrt{3}) \;\Rightarrow\; \text{no inverse on } \mathbb{R}f(0)=0=f(3)⇒no inverse on R
On an interval where fff stays monotonic it is one-to-one and does have an inverse; the reciprocal is unrelated to inverting.
fff is one-to-one and f(−3)=5f(-3) = 5f(−3)=5. Which point is on the graph of f−1f^{-1}f−1?
Inverting swaps the coordinates of the pair.
f(−3)=5 ⇒ f−1(5)=−3 ⇒ (5,−3) on f−1f(-3) = 5 \;\Rightarrow\; f^{-1}(5) = -3 \;\Rightarrow\; (5, -3) \text{ on } f^{-1}f(−3)=5⇒f−1(5)=−3⇒(5,−3) on f−1
For f(x)=x2f(x) = x^2f(x)=x2 with domain x≥0x \ge 0x≥0 and g(x)=xg(x) = \sqrt{x}g(x)=x, both compositions give xxx. Why is this different from the all-reals case?
On all reals g(f(x))=x2=∣x∣g(f(x)) = \sqrt{x^2} = |x|g(f(x))=x2=∣x∣ fails for x<0x < 0x<0; the restriction removes those inputs.
x≥0 ⇒ x2=∣x∣=xx \ge 0 \;\Rightarrow\; \sqrt{x^2} = |x| = xx≥0⇒x2=∣x∣=x
Same formulas, smaller domain, and now both compositions are the identity, so ggg genuinely inverts fff.
Which statement about a one-to-one function fff and its inverse is FALSE?
The first three are always true; the last is the classic overclaim.
f(x)=−x3 meets f−1 at (1,−1) and (−1,1)f(x) = -x^3 \text{ meets } f^{-1} \text{ at } (1, -1) \text{ and } (-1, 1)f(x)=−x3 meets f−1 at (1,−1) and (−1,1)
Those crossings are off y=xy = xy=x, so 'meet only on y=xy = xy=x' is false.
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