AP Precalc 2.7-2.9 Visual Study Guide: Compositions, Inverses & LogarithmsAP Precalc 2.7-2.9 Visual Study Guide: Compositions, Inverses & Logarithms

A P P R E C A L C U L U S · UNIT2B · C O N C E P T C H E C K 2 . 7 – 2 . 9

Compositions, Inverses & Logarithms

Study guide for Topics 2.7 (Composition of Functions), 2.8 (Inverse Functions) and 2.9 (Logarithmic Expressions). Built from the Unit 2B notes packet, the 2.7 and 2.8 notes keys and the 2.7 and 2.8 homework keys.

WHAT THE CONCEPT CHECK ASKS YOU TO DO

Evaluate compositions from equations, tables, graphs, piecewise rules and verbal descriptions, including "if possible" ones that come out undefined.

Build composite rules like f(g(x)), state their domain, and decompose a function into an inside and an outside piece.

Find inverses from a table, a graph and an equation, including rational functions with x in the top and bottom.

Verify two functions are inverses by showing both compositions equal x.

Use inverse properties: swap domain and range, read f⁻¹ values from a table of f, and justify invertibility in words.

Convert between exponential and log form, evaluate logs without a calculator, and read log-scaled graphs.

The one idea that ties the three topics together

All three lessons are about running functions in a chain and backwards.

x → g → g(x) → f → f(g(x))

Topic The move One-line version

2.7

Composition

Chain two functions: one output becomes the next

input.

(f∘g)(x) = f(g(x)). Work from the inside

out.

2.8 Inverses Run a function backwards: switch every input and

output.

f⁻¹(f(x)) = x. Domain and range trade

places.

2.9 Logarithms The inverse of exponentiation. ba = c ⇔ logbc = a. A log is an exponent. Topic 2.9 is really 2.8 applied to one function. The log asks "what exponent?", and that question undoes bx. Keep that link in mind and 2.10 and 2.11 will feel like review.

2.7 · Composition of Functions

20-SECOND SUMMARY

(f∘g)(x) = f(g(x)), read "f composed with g at x" or "f of g of x". Do g first, then feed that output into f. Order matters: f(g(x)) and g(f(x)) are usually different functions. The composite only exists where x is in the domain of g and g(x) is in the domain of f.

Reading the notation

The small open circle ∘ is the composition operation. It is not multiplication: f(g(x)) ≠ f(x)·g(x). The function written closest to x (the rightmost one) acts first.

Written Means Order of work

(f∘g)(3) f(g(3)) g, then f

(g∘f)(3) g(f(3)) f, then g

(f∘f)(4) f(f(4)) f, then f again

(g∘f∘h∘k)(3) g(f(h(k(3)))) k, h, f, g (right to left)

Evaluating at a number (EX #2 from the notes)

f(x) = 3x − 5 and g(x) = 2x + 1.

f(g(3))

g(3) = 2(3) + 1 = 7

f(7) = 3(7) − 5 = 16

g(f(3))

f(3) = 3(3) − 5 = 4

g(4) = 2(4) + 1 = 9

Same two functions, same input, different answers. That is the whole reason order matters.

TRAP

Doing the outside function first. For f(g(3)), students plug 3 into f (getting 4) and then into g (getting 9). That is g(f(3)). Box the inner value first, the way the notes key does, then move outward. Building a composite rule (EX #3)

Replace every x in the outer function with the whole inner expression, in parentheses.

f(g(x)) = f(2x + 1)

= 3(2x + 1) − 5

= 6x + 3 − 5 = 6x − 2

g(f(x)) = g(3x − 5)

= 2(3x − 5) + 1

= 6x − 10 + 1 = 6x − 9

From the homework, with f(x) = 4x − 5 and g(x) = x² − 2x + 4: f(g(x)) = 4(x² − 2x + 4) − 5 = 4x² − 8x + 11. Note the 4 distributes to every term of the inner quadratic.

Exponential inside or outside works the same way. With h(x) = 3(2)x and k(x) = 3 − 2x:

h(k(x)) = 3(2)3−2x = 3 · 2³ · (2−2)x = 24( 1

4 )x

That rewrite uses the exponent rules from 2.4 and shows the composite is a decreasing exponential (base 1/4).

TRAP

Squaring a binomial wrong inside a composition. If f(x) = x² and g(x) = x + 3, then f(g(x)) = (x + 3)² = x² + 6x + 9, not x² + 9.

QUICK CHECK

With f(x) = 4x − 5 and k(x) = 3 − 2x, find k(f(x)).

Answer: 3 − 2(4x − 5) = 3 − 8x + 10 = −8x + 13

Domain of a composite (EX #4)

Two filters, in order: (1) x must go into the inner function; (2) the inner output must go into the outer function. Find the domain from the composition steps, not just from the simplified final expression. f(x) = x² − 3x, g(x) = 1

x + 2 , h(x) = √x.

(g∘f)(x) = g(x² − 3x) = 1

x² − 3x + 2 = 1

(x − 2)(x − 1)

Denominator ≠ 0, so x ≠ 1, 2.

D: (−∞, 1) ∪ (1, 2) ∪ (2, ∞)

(h∘g)(x) = h( 1

x + 2 ) = √( 1

x + 2 ) = 1

√(x + 2)

Need x + 2 > 0 (positive under the root and not zero

in the bottom).

D: (−2, ∞)

From EX #5: (k∘p)(x) with p(x) = √(3x + 1) and k(x) = x² + 4x + 5 simplifies to 3x + 6 + 4√(3x + 1), with domain [−1/3, ∞) because 3x + 1 ≥ 0.

TRAP

Reading the domain off the simplified answer. If f(x) = x² and g(x) = √x, then f(g(x)) = (√x)² = x. The simplified rule looks like it works for every real number, but g(−4) is undefined, so the composite's domain is x ≥ 0.

Compositions from tables, graphs, piecewise rules and words (EX #7)

This is the most-tested 2.7 skill on the AP-style concept check. The functions come in different representations, and you hop between them.

Graph of f (EX #7 style)

-5 -4 -3 -2 -1123456789 10 11

-3

-2

-1

1

2

3

4

5fx −41257

g(x) π 3 3 −2 4

h(x) = |−7 + 3x| if x < 1; −4 if x = 1; −x if x > 1

k is a quadratic: y = x² shifted right 2, then down 3,

so k(x) = (x − 2)² − 3.

Find Work (inside out) Answer

f(g(5)) g(5) = −2 from the table; f(−2) = 2 from the graph 2

(h∘f)(8) f(8) = −1 from the graph; −1 < 1, so use the first piece: h(−1) = |−7 + 3(−1)| = | −10|

10

f(f(2)) f(2) = 1 (closed dot at (2, 1)); f(1) = 4 4

(g∘k)(0) k(0) = (0 − 2)² − 3 = 1; g(1) = 3 3

(g∘h)(1) h(1) = −4 (the "x = 1" piece); g(−4) = π π

(g∘f∘h∘k)

(3)

k(3) = −2; h(−2) = |−7 − 6| = 13; f(13) is off the graph Does not

exist

TRAP

Choosing the piece by the output instead of the input. In h(f(8)), you pick h's piece using the number you are plugging into h (which is −1), not using 8.

TRAP

Forcing an answer when the chain leaves the domain. If any inner output lands outside the next function's domain (off the end of a graph, a missing table value, a zero denominator), the composite does not exist there. "Evaluate if possible" is a hint that one of them won't be.

TRAP