brought the first real resistance. The function ( g(x) = \frac{3x+1}{x-2} ), ( x \neq 2 ). Find ( g^{-1}(x) ) and state its domain. She swapped ( x ) and ( y ): ( x = \frac{3y+1}{y-2} ). Cross-multiplied: ( x(y-2) = 3y+1 ). ( xy - 2x = 3y + 1 ). Grouped terms: ( xy - 3y = 2x + 1 ). Factored: ( y(x-3) = 2x+1 ). So ( g^{-1}(x) = \frac{2x+1}{x-3} ).
hit her like a cold splash of water. Given that ( f(x) = 2x^3 + 3x^2 - 8x + 3 ), show that ( (x-1) ) is a factor, and hence fully factorise ( f(x) ). Elena took a breath. Polynomials. I can do this. She scribbled the substitution: ( f(1) = 2 + 3 - 8 + 3 = 0 ). Yes. Then came the algebraic long division, the careful subtraction of terms, the descent into the quadratic. ( (x-1)(2x^2 + 5x - 3) ). Then the final break: ( (x-1)(2x-1)(x+3) ). core pure -as year 1- unit test 5 algebra and functions
On her desk lay . The front cover was deceptively calm, featuring only the exam board’s logo and the instruction: Attempt all questions. Use algebraic methods unless otherwise stated. brought the first real resistance
Roots: ( x = 2 ) and ( x = -2 ), both repeated (multiplicity 2). The inequality ( p(x) < 0 ) asked: when is a square less than zero? She swapped ( x ) and ( y ): ( x = \frac{3y+1}{y-2} )
And for the first time, she felt like a real mathematician.
She turned the page.
The answer formed: ( \frac{1}{x-1} - \frac{1}{x+2} + \frac{5}{x-3} ). Clean. Elegant.