Find domain and range of Q f (x) = x2+2x+1x2-8x+12 - Maths - Relations and Functions

Question

Find domain and range of
Q f (x) = x 2 + 2 x + 1 x 2 - 8 x + 12

Aarushi Mishra · Accepted Answer

fx=x2+2x+1x2-8x+12fx=x2+2x+1x2-6x-2x+12fx=x2+2x+1xx-6-2x-6fx=x2+2x+1x-2x-6Since denominator should not be zero therefore, x≠2,6
 Rest for all x this will be definedSo domain=ℝ-2,6Let y=x2+2x+1x2-8x+12yx2-8yx+12y=x2+2x+1y-1x2-8y+2x+12y-1=0Since x∈ℝ for  this equation to have real roots
discriminant, D≥0D=b2-4ac=-8y+22-4y-112y-1≥044y+12-4y-112y-1≥04y+12-y-112y-1≥016y2+8y+1-12y2-13y+1≥016y2+8y+1-12y2+13y-1≥04y2+21y≥04yy+214≥0y-0y--214≥0We know that if y-ay-b≥0 , where a<b then x∈(-∞,a]∪[b,∞)So, herey∈(-∞,0]∪[-214,∞)Therefore range=(-∞,0]∪[214,∞)

Find domain and range of Q. f (x) = x 2 + 2 x + 1 x 2 - 8 x + 12

Find domain and range of
Q. f (x) = $\frac{x^{2} + 2 x + 1}{x^{2} - 8 x + 12}$