Functions
A function is a rule with one promise attached: give it an input and it returns exactly one output. Everything here follows from that promise. The last section, the difference quotient, is the one that matters most later — it is the algebra a derivative is built out of, so the fluency you build on it now is spent in every chapter that follows.
Calculus — graphing and functions
Graphing calculator- The promiseone input ↦ exactly one output
Two outputs for one input and it is not a function. Two inputs sharing one output is perfectly fine.
- Function notationf(x) = 3x − 1
f(x) is not f times x. It names the output that the rule f gives for the input x.
- Evaluatingf(x) = 3x − 1 ⇒ f(5) = 3(5) − 1
Whatever goes in the brackets goes everywhere the x was, and goes in bracketed. That habit is what stops a sign, or a whole expression, breaking loose later.
- Vertical line testa vertical line may cross the graph at most once
Two crossings would be one input with two outputs, which the promise forbids.
- Composition(f ∘ g)(x) = f(g(x))
Inside out: g acts first. The order matters and swapping it usually gives a different function.
- Difference quotientf(x + h) − f(x)h
The slope between two points on the graph. Simplify until the h in the denominator cancels — that cancellation is the point of the exercise.
- 1
Start with every real number. A polynomial never breaks, so if there is no fraction, no even root and no log, the domain is all reals and there is nothing to do.
- 2
Throw out anything that makes a denominator zero. Set each denominator equal to zero and solve; those values are excluded. 1x−4 loses x = 4.
- 3
Throw out anything that makes an even root negative. Set the inside ≥ 0 and solve. √2x − 6 needs x ≥ 3. An odd root takes anything, so a cube root adds no restriction at all.
- 4
Both at once means strictly greater. An even root in a denominator has to be positive rather than merely non-negative, because zero would divide by zero. That turns a ≥ into a >.
- 5
Write it as intervals. Round brackets for an excluded end, square for an included one, and ∪ to join the pieces. Infinity is always excluded.
- Sum and difference(f ± g)(x) = f(x) ± g(x)
Work out each output separately, then combine the two numbers. Nothing clever happens.
- Product(f · g)(x) = f(x) · g(x)
Two radicals multiplied can be pulled under one radical sign, which tidies the answer and hides the domain. Write the domain down before you tidy.
- Quotientfg(x) = f(x)g(x), g(x) ≠ 0
The quotient carries one restriction the others do not: every x with g(x) = 0 is thrown out, even though it sat happily inside both original domains.
- Domain of any combinationdomain of f ∩ domain of g
Both rules have to be computable before they can be combined, so only the overlap survives. The stricter of the two conditions wins.
The domain comes from the two functions you started with, never from the expression you finish with.
- A rule with two branchesf(x) = 1 − x when x ≤ −1
The condition is as much a part of the rule as the formula is. Reading the formula without it is the usual mistake.
- ...and the other branchf(x) = x2 when x > −1
The conditions must cover the domain without overlapping. Exactly one branch owns any given input, which is what keeps it a function.
- Absolute value is piecewise|x| = −x when x < 0, x when x ≥ 0
It returns the distance from zero, so it is never negative. The minus sign on the first branch is what turns a negative input positive.
- Filled and hollow endpointsfilled dot included, hollow dot excluded
At a boundary the branch whose condition uses ≤ or ≥ takes the filled dot; the branch with a strict < or > takes the hollow one.
- Step functionconstant across an interval, then a jump
A piecewise function whose every branch is a constant. The graph is a run of flat segments, each with one filled and one hollow end.
- f(x + h) = f(x) + hf(x + h) means: put x + h in for every x
Function notation is not multiplication and does not distribute. For f(x) = x2: f(3 + 1) = 16, while f(3) + 1 = 10.
- f(x)2 = f(x2)f(x)2 squares the output; f(x2) squares the input
For f(x) = x + 1: f(3)2 = 16, while f(32) = 10. Two different instructions.
- x + 3x2 − 9 = 1x − 3 , so the domain is x ≠ 3The domain is still x ≠ −3 and x ≠ 3
Cancelling simplifies the expression, but the function was defined by the original rule and that rule was never given a value at −3. Simplifying cannot hand it one.
- domain of (f · g) , read off √x2 − 3x − 4domain of f ∩ domain of g
Combining two rules can tidy into an expression that would accept more inputs than either original did. √x + 1√x − 4 becomes √x2 − 3x − 4, which alone would take x = −2 \u2014 but g(−2) = √−6 never existed.
- at a boundary, use whichever branch is easieruse the branch whose condition includes the boundary
Exactly one branch may own the boundary. Against x ≤ −1 and x > −1, the input x = −1 is the top rule's. Handing it to both would put two outputs on one input.
- f(x) = x2 is not a function, because f(2) and f(−2) both give 4It is a function.
The promise is one output per input, not one input per output. Two inputs sharing an output is ordinary — it only means the graph fails the horizontal line test, which is about invertibility, not about being a function.
Type an answer and it gets marked. What you get right is kept on this device.
- 1.f(x) = 4 − x2, f(−3)
- 2.f(x) = 3x + 1, f(x + h) − f(x)h
- 3.f(x) = 5x2 − x − 6, domain
- 4.g(x) = √9 − 3x, domain
- 5.f(x) = x − 4, g(x) = x2, (f ∘ g)(x)
- 6.f(x) = 2x − 3, g(x) = x2, fg(x) and its domain
- 7.f(x) = √x, g(x) = x2 − 1, (f ∘ g)(x) and its domain
- 8.f(x) = x + 4 when x < −2, f(x) = x2 when x ≥ −2, f(−3) and f(−2)
- 9.f(x) = x2 + x, f(x + h) − f(x)h