f(x) = 3x − 2 and g(x) = x2 + 1 for all real x. Solve gf(x) = 17.
- Apply f first: gf(x) = g(3x − 2) = (3x − 2)2 + 1.
- (3x − 2)2 + 1 = 17, so (3x − 2)2 = 16 and 3x − 2 = 4 or −4.
- 3x = 6 gives x = 2. 3x = −2 gives x = −2/3.
Answer: x = 2 or x = −2/3
AS Level Mathematics · Pure Mathematics 1. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.
A function is a rule that gives exactly one output for each input. The domain is the set of inputs allowed. The range is the set of outputs that the function actually produces. To find a range, sketch the graph over the given domain, or use the completed square form for a quadratic.
A function is one-one if every output comes from only one input. Only a one-one function has an inverse. The inverse f−1 undoes f, so the domain and range swap over.
A composite function applies one function and then another. gf can only be formed when the range of f lies inside the domain of g. Graphs can also be moved or stretched by changing the formula in simple ways.
Students read fg(x) from left to right and apply f first. The function nearest to x acts first, so fg(x) means g and then f. Also, f(x + a) moves the graph to the left, not to the right.
Answer: x = 2 or x = −2/3
Try each one, then open the answer.
Answer: C. Complete the square: f(x) = (x − 2)2 − 3, vertex at x = 2. Restricting the domain to x ≥ 2 keeps only one arm of the parabola, so f is one-one; smallest k = 2.
Answer: D. Inverting swaps the roles of x and y, so every point (a, b) on y = f(x) becomes (b, a) on y = f−1(x); this is reflection in the line y = x.
Answer: B. y = 4 − (x + 1)2 gives (x + 1)2 = 4 − y, and since x ≤ −1 we need x + 1 ≤ 0, so x + 1 = −√(4 − y). Hence f−1(x) = −1 − √(4 − x); its domain is the range of f, which is x ≤ 4.
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