IB AA · 3A Concepts of a Function
Find every output the rule can produce.
Quadratic
The axis is x = −b2a. Substitute that x to find the turning value.
If a > 0, the range is y ≥ that value. If a < 0, the range is y ≤ that value.
Restricted line
Look at the gradient. The endpoint of the domain gives the endpoint of the range. An open end of the domain stays open in the range.
Square root
A square root is ≥ 0. Multiply by a negative coefficient and the inequality reverses.
Logarithm
On a restricted domain, evaluate the endpoint, then follow the log as x grows.
A quadratic for all real x.
f(x) = x² − 6x + 5
x = −b2a
x = 62
x = 3
f(3) = (3)² − 6(3) + 5
f(3) = −4
a is positive, so the smallest output is −4.
The range is y ≥ −4.
Answer: y ≥ −4
f(x) = x² − 3
x = −b2a
x = 02
x = 0
f(0) = (0)² − 3
f(0) = −3
a is positive, so the smallest output is −3.
The range is y ≥ −3.
Answer: y ≥ −3
f(x) = x² − 4x + 1
x = −b2a
x = 42
x = 2
f(2) = (2)² − 4(2) + 1
f(2) = −3
a is positive, so the smallest output is −3.
The range is y ≥ −3.
Answer: y ≥ −3
f(x) = −x² + 6
x = −b2a
x = 0−2
x = 0
f(0) = −(0)² + 6
f(0) = 6
a is negative, so the largest output is 6.
The range is y ≤ 6.
Answer: y ≤ 6
f(x) = x² + 2x + 4
x = −b2a
x = −22
x = −1
f(−1) = (−1)² + 2(−1) + 4
f(−1) = 3
a is positive, so the smallest output is 3.
The range is y ≥ 3.
Answer: y ≥ 3
f(x) = −x² + 6x − 5
x = −b2a
x = −6−2
x = 3
f(3) = −(3)² + 6(3) − 5
f(3) = 4
a is negative, so the largest output is 4.
The range is y ≤ 4.
Answer: y ≤ 4
A restricted line, or a square root.
f(x) = 2x² − 8x + 3
x = −b2a
x = 84
x = 2
f(2) = 2(2)² − 8(2) + 3
f(2) = −5
a is positive, so the smallest output is −5.
The range is y ≥ −5.
Answer: y ≥ −5
f(x) = 2x + 8, x ≤ 5
The gradient is positive, so the largest output is at the right endpoint.
f(5) = 2(5) + 8
f(5) = 18
The range is y ≤ 18.
Answer: y ≤ 18
f(x) = −3x + 7, x > 1
The gradient is negative. Just to the right of the endpoint, y is just under the endpoint value, and y then falls without a limit.
f(1) = −3(1) + 7
f(1) = 4
The range is y < 4.
Answer: y < 4
f(x) =
The square root is at least 0, and the coefficient is positive, so every output is at least 0. The root grows without a limit.
The range is y ≥ 0.
Answer: y ≥ 0
f(x) = −
The square root is at least 0. Multiplying by a negative number makes every output at most 0, with no lower bound.
The range is y ≤ 0.
Answer: y ≤ 0
f(x) = 3x + 1, x ≤ 4
The gradient is positive, so the largest output is at the right endpoint.
f(4) = 3(4) + 1
f(4) = 13
The range is y ≤ 13.
Answer: y ≤ 13
A logarithm, or a square root subtracted from a number.
f(x) = 2 + ln(x − 7), and x ≥ 8
At the endpoint, x = 8.
f(8) = 2 + ln(1)
ln 1 = 0, so f(8) = 2
The log increases as x increases, so the outputs grow without a limit.
The range is y ≥ 2.
Answer: y ≥ 2
g(x) = 4 − ln(2x + 1), and x > 0
As x approaches 0 from the right, 2x + 1 approaches 1.
ln 1 = 0, so g(x) approaches 4, but x is never 0, so g(x) never reaches 4.
As x grows, the log grows, so subtracting it sends g(x) down without a limit.
The range is y < 4.
Answer: y < 4
f(x) = x² + 3x + 1
x = −b2a
x = −32
x = −32
f(−32) = (−32)² + 3(−32) + 1
f(−32) = −54
a is positive, so the smallest output is −54.
The range is y ≥ −54.
Answer: y ≥ −54
f(x) = 4x − 2, x ≥ −1
The gradient is positive, so the smallest output is at the left endpoint.
f(−1) = 4(−1) − 2
f(−1) = −6
The range is y ≥ −6.
Answer: y ≥ −6
f(x) = 5 −
A square root is at least 0, so subtracting it gives a result ≤ 5.
The root can grow without a limit, so f(x) has no lower bound.
The range is y ≤ 5.
Answer: y ≤ 5
f(x) = −2x² + 4x + 3
x = −b2a
x = −4−4
x = 1
f(1) = −2(1)² + 4(1) + 3
f(1) = 5
a is negative, so the largest output is 5.
The range is y ≤ 5.
Answer: y ≤ 5
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