AP Precalculus Flashcards: Inverse Functions

Study Inverse Functions in AP Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Precalculus

Inverse Functions

0 mastered0 still learning

0% Complete

QUESTION
1/ 39

What is the inverse of the logarithmic function f(x)=loga(x)f(x)=\log_a(x) for a>0a>0 and a1a\ne 1?

Tap card or press Space to flip

ANSWER

f1(x)=axf^{-1}(x)=a^x. Exponentials and logarithms are inverse operations.

How well did you know it?

Card 1 / 39

What this deck covers

This deck focuses on Inverse Functions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

All flashcards

Flashcard 1: What is the inverse of the logarithmic function f(x)=loga(x)f(x)=\log_a(x) for a>0a>0 and a1a\ne 1?

Answer: f1(x)=axf^{-1}(x)=a^x. Exponentials and logarithms are inverse operations.

Flashcard 2: What is the standard algebraic procedure to find f1(x)f^{-1}(x) from y=f(x)y=f(x)?

Answer: Swap xx and yy, then solve for yy and rename yy as f1(x)f^{-1}(x). This process reverses the input-output relationship of ff.

Flashcard 3: Identify the inverse of f(x)=ln(x)f(x)=\ln(x).

Answer: f1(x)=exf^{-1}(x)=e^x. Natural exponential is the inverse of natural log.

Flashcard 4: Identify the inverse of f(x)=exf(x)=e^x.

Answer: f1(x)=ln(x)f^{-1}(x)=\ln(x). Natural log is the inverse of the natural exponential.

Flashcard 5: Identify a domain restriction that makes f(x)=x2f(x)=x^2 have an inverse function.

Answer: Restrict to [0,)[0,\infty) or to (,0](-\infty,0]. On these intervals, f(x)=x2f(x)=x^2 is strictly monotonic.

Flashcard 6: Identify the inverse of f(x)=(x2)3+1f(x)=(x-2)^3+1.

Answer: f1(x)=x13+2f^{-1}(x)=\sqrt[3]{x-1}+2. Undo operations in reverse order: subtract 1, take cube root, add 2.

Flashcard 7: Find f1(x)f^{-1}(x) for f(x)=3x5f(x)=3x-5.

Answer: f1(x)=x+53f^{-1}(x)=\frac{x+5}{3}. Swap: x=3y5x=3y-5, solve: y=x+53y=\frac{x+5}{3}.

Flashcard 8: Identify the inverse function of f(x)=3x5f(x)=3x-5.

Answer: f1(x)=x+53f^{-1}(x)=\frac{x+5}{3}. Swap xx and yy, then solve for yy.

Flashcard 9: What graphical transformation relates the graphs of y=f(x)y=f(x) and y=f1(x)y=f^{-1}(x)?

Answer: Reflection across the line y=xy=x. Points (a,b)(a,b) and (b,a)(b,a) are reflections across this line.

Flashcard 10: What is the inverse of the point (a,b)(a,b) on y=f(x)y=f(x) as a point on y=f1(x)y=f^{-1}(x)?

Answer: (b,a)(b,a). Swapping coordinates reflects the point across y=xy=x.

Flashcard 11: What is the definition of an inverse function in terms of composition?

Answer: f1(f(x))=xf^{-1}(f(x))=x and f(f1(x))=xf(f^{-1}(x))=x on the appropriate domains. These compositions yield the identity function on their respective domains.

Flashcard 12: What is the correct meaning of the notation f1(x)f^{-1}(x)?

Answer: It is the inverse function of ff, not the reciprocal 1f(x)\frac{1}{f(x)}. The 1-1 is an exponent notation, not a negative power.

Flashcard 13: What condition must a function satisfy to have an inverse function that is also a function?

Answer: It must be one-to-one (pass the horizontal line test). Each output must correspond to exactly one input.

Flashcard 14: What is the Horizontal Line Test used to determine about a function ff?

Answer: Whether ff is one-to-one (and therefore invertible on its domain). If any horizontal line hits the graph twice, ff has no inverse.

Flashcard 15: Identify whether f(x)=x2f(x)=x^2 is invertible on (,)(-\infty,\infty).

Answer: Not invertible on (,)(-\infty,\infty) (not one-to-one). Both x=2x=2 and x=2x=-2 give f(x)=4f(x)=4, failing one-to-one.

Flashcard 16: What does it mean for a function ff to have an inverse function f1f^{-1}?

Answer: ff is one-to-one; each yy in the range comes from exactly one xx. One-to-one means no horizontal line intersects the graph more than once.

Flashcard 17: Which restriction makes f(x)=x2f(x)=x^2 invertible as a function?

Answer: Restrict the domain to x0x\ge 0 (or to x0x\le 0). This makes the function pass the horizontal line test.

Flashcard 18: What is the inverse of the exponential function f(x)=axf(x)=a^x for a>0a>0 and a1a\ne 1?

Answer: f1(x)=loga(x)f^{-1}(x)=\log_a(x). Logarithms and exponentials are inverse operations.

Flashcard 19: Evaluate f1(7)f^{-1}(7) for f(x)=2x+1f(x)=2x+1.

Answer: f1(7)=3f^{-1}(7)=3. Since f(3)=7f(3)=7, we have f1(7)=3f^{-1}(7)=3.

Flashcard 20: Identify the inverse relation of the equation y=2x+35y=\frac{2x+3}{5}.

Answer: y=5x32y=\frac{5x-3}{2}. Swap xx and yy: x=2y+35x=\frac{2y+3}{5}, solve for yy.

Flashcard 21: Find and correct the error: claiming f1(x)=1f(x)f^{-1}(x)=\frac{1}{f(x)} for an inverse function.

Answer: Correct: f1f^{-1} is not a reciprocal; it satisfies f(f1(x))=xf(f^{-1}(x))=x. The inverse function undoes ff, not reciprocates it.

Flashcard 22: Given f(2)=9f(2)=9, what is f1(9)f^{-1}(9)?

Answer: f1(9)=2f^{-1}(9)=2. By definition, f1f^{-1} undoes ff: if f(a)=bf(a)=b, then f1(b)=af^{-1}(b)=a.

Flashcard 23: What is the inverse of the linear function f(x)=mx+bf(x)=mx+b with m0m\ne 0?

Answer: f1(x)=xbmf^{-1}(x)=\frac{x-b}{m}. Solve y=mx+by=mx+b for xx to get the inverse formula.

Flashcard 24: What happens to a point (a,b)(a,b) on y=f(x)y=f(x) when graphed on y=f1(x)y=f^{-1}(x)?

Answer: It becomes the point (b,a)(b,a). Swapping coordinates reflects the point across y=xy=x.

Flashcard 25: Identify the inverse function of f(x)=x47f(x)=\frac{x-4}{7}.

Answer: f1(x)=7x+4f^{-1}(x)=7x+4. Multiply by 7 and add 4 to undo the original operations.

Flashcard 26: What is the inverse of the exponential function f(x)=bxf(x)=b^x for b>0b>0 and b1b\ne 1?

Answer: f1(x)=logb(x)f^{-1}(x)=\log_b(x). Logarithms and exponentials are inverse operations.

Flashcard 27: What is the inverse of the linear function f(x)=mx+bf(x)=mx+b with m0m\ne 0?

Answer: f1(x)=xbmf^{-1}(x)=\frac{x-b}{m}. Solve y=mx+by=mx+b for xx to get x=ybmx=\frac{y-b}{m}.

Flashcard 28: What is the relationship between the domain and range of ff and f1f^{-1}?

Answer: Dom(f1)=Range(f)\text{Dom}(f^{-1})=\text{Range}(f) and Range(f1)=Dom(f)\text{Range}(f^{-1})=\text{Dom}(f). The inverse swaps the input and output sets of ff.

Flashcard 29: What is the key graph relationship between y=f(x)y=f(x) and y=f1(x)y=f^{-1}(x)?

Answer: They are reflections across the line y=xy=x. Points (a,b)(a,b) and (b,a)(b,a) are symmetric about y=xy=x.

Flashcard 30: Identify whether f(x)=x2f(x)=x^2 is one-to-one on the domain (,)(-\infty,\infty).

Answer: No, it is not one-to-one on (,)(-\infty,\infty). It fails the horizontal line test (e.g., f(2)=f(2)=4f(-2)=f(2)=4).

Flashcard 31: Find f1(x)f^{-1}(x) for f(x)=x42f(x)=\frac{x-4}{2}.

Answer: f1(x)=2x+4f^{-1}(x)=2x+4. Swap: x=y42x=\frac{y-4}{2}, solve: y=2x+4y=2x+4.

Flashcard 32: Find f1(x)f^{-1}(x) for f(x)=x1f(x)=\sqrt{x-1} with domain x1x\ge 1.

Answer: f1(x)=x2+1f^{-1}(x)=x^2+1 with domain x0x\ge 0. Swap: x=y1x=\sqrt{y-1}, square both sides: y=x2+1y=x^2+1.

Flashcard 33: What is the inverse of f(x)=xf(x)=\sqrt{x} when the domain is x0x\ge 0?

Answer: f1(x)=x2f^{-1}(x)=x^2 with domain x0x\ge 0. Squaring undoes the square root for non-negative values.

Flashcard 34: What is the inverse of the identity function f(x)=xf(x)=x?

Answer: f1(x)=xf^{-1}(x)=x. The identity function is its own inverse since f(f(x))=xf(f(x))=x.

Flashcard 35: What is the inverse of f(x)=x3f(x)=x^3?

Answer: f1(x)=x3f^{-1}(x)=\sqrt[3]{x}. The cube root undoes the cubing operation.

Flashcard 36: What is the composition value f(f1(x))f(f^{-1}(x)) for inputs xx in the domain of f1f^{-1}?

Answer: f(f1(x))=xf(f^{-1}(x))=x. By definition of inverse functions.

Flashcard 37: What is the relationship between the domain and range of ff and f1f^{-1}?

Answer: Dom(f1)=Ran(f)\text{Dom}(f^{-1})=\text{Ran}(f) and Ran(f1)=Dom(f)\text{Ran}(f^{-1})=\text{Dom}(f). The domain and range swap when finding the inverse.

Flashcard 38: What is the defining composition property of inverse functions ff and f1f^{-1}?

Answer: f(f1(x))=xf(f^{-1}(x))=x and f1(f(x))=xf^{-1}(f(x))=x (on appropriate domains). These compositions yield the identity function on their respective domains.

Flashcard 39: Find f1(x)f^{-1}(x) for f(x)=(x+2)3f(x)=(x+2)^3.

Answer: f1(x)=x32f^{-1}(x)=\sqrt[3]{x}-2. Swap: x=(y+2)3x=(y+2)^3, take cube root: y=x32y=\sqrt[3]{x}-2.