Function Notation and Evaluating Functions
Read f(x) as a rule that turns each input into one output, evaluate a function at given values, and solve f(x) = k.
What a learner can do afterwards
- Work out f(3) and f(-2) for a given rule
- Solve f(x) = 0 for a quadratic rule
- Explain why each input is allowed only one output
1 · Read
You have been doing this all along without the name. The nth term rule n squared + 2n + 1 takes a position number in and gives a term out. That is a function: a rule that turns each input into exactly one output. Function notation just puts a label on the rule. Instead of saying the rule is 3x - 2, we write f(x) = 3x - 2 and call the rule f. The x in the brackets is a placeholder, the slot where the input goes. Writing f(4) means feed 4 into the rule and report what comes out.
Evaluating a function is substitution with better manners. To find f(-1) for f(x) = 3x - 2, put -1 in every place the x appears: 3 times -1 is -3, minus 2 is -5. Watch the brackets with negative inputs, because 3(-1) is -3, not 3 - 1. And the one-output rule is real: each input is allowed exactly one output, because the rule gives no choice about what to do with it. Two different inputs may share an output, like 3 and -3 both landing on 5 in x squared - 4, but one input never splits into two outputs.
Solving f(x) = k runs the machine in reverse. You are not given the input, you are given the output k, and you must find the input or inputs that produce it. Set the rule equal to k and solve the resulting equation. If k is zero, you are just finding the roots of the rule, which you have done many times. A linear rule gives one input for each k. A quadratic rule can give two, or one when the curve only touches, or none at all when the output k lies on the wrong side of the turning point.
Evaluate f(4), f(-1) and f(0) for f(x) = 3x - 2. At 4: 3 times 4 is 12, minus 2 is 10. At -1: 3 times -1 is -3, minus 2 is -5. The brackets keep the sign honest. At 0: 3 times 0 is 0, minus 2 is -2. Notice that f(0) just hands you the constant term. That is a useful check when you are writing rules backwards, because the constant is always the output at zero.
Now a rule that bends. Evaluate g(3), g(-3) and g(2) for g(x) = x squared - 4. At 3: 9 - 4 is 5. At -3: (-3) squared is still 9, so 9 - 4 is 5 again. At 2: 4 - 4 is 0. So the inputs 3 and -3 share the output 5, which is allowed. The rule still gave each input exactly one answer. But g(-3) is not -9 - 4, because squaring comes before the minus sign is applied to the result, and a negative number squared is positive. That one slip is the most common error on this topic.
Finally, run it in reverse. For h(x) = x squared - 5x + 6, find which inputs give output 0. Set the rule to zero: x squared - 5x + 6 = 0. The pair that multiplies to 6 and adds to -5 is -2 and -3, so it factors as (x - 2)(x - 3), and the inputs are 2 and 3. Check: h(2) is 4 - 10 + 6 = 0. Now the output 2: subtract 2 to get x squared - 5x + 4 = 0, which factors as (x - 1)(x - 4): inputs 1 and 4. Same method throughout.
Read f(x) as a rule with a placeholder, evaluate by substituting the input with care for negative numbers, and solve f(x) = k by setting the rule equal to k and solving for the inputs.
2 · Watch
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Where this leads
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24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.