Roots: ( x = 2 ) and ( x = -2 ), both repeated (multiplicity 2). The inequality ( p(x) < 0 ) asked: when is a square less than zero?
was the function composition trap. Given ( h(x) = \sqrt{x+4} ) for ( x \geq -4 ), and ( k(x) = x^2 - 1 ) for ( x \geq 0 ). Find ( h(k(x)) ) and state its domain. She composed carefully: ( h(k(x)) = \sqrt{(x^2 - 1) + 4} = \sqrt{x^2 + 3} ). Wait, she thought. That’s defined for all real ( x ), but ( k ) only takes ( x \geq 0 ). And ( k(x) ) gives outputs ( \geq -1 ), but ( h ) requires inputs ( \geq -4 ). That’s fine. core pure -as year 1- unit test 5 algebra and functions
The answer formed: ( \frac{1}{x-1} - \frac{1}{x+2} + \frac{5}{x-3} ). Clean. Elegant. Roots: ( x = 2 ) and (
She turned the page.
Elena set her pen on the desk. Her palms were damp, but her mind was clear. She had faced the domain restrictions, the partial fraction decomposition, the inverse function trap, the composite’s hidden conditions, and the elegant emptiness of the squared inequality. Given ( h(x) = \sqrt{x+4} ) for (
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} ).