Consider f: R+ â†’ [âˆ’5, âˆž) given by f(x) = 9x2 + 6x âˆ’ 5 Show that f is - Maths - Relations and Functions

Question

Consider f: R+ â†’ [âˆ’5,
âˆž) given by f(x) = 9x2 + 6x
âˆ’ 5 Show that f is one one,onto hence
invertible

Ajanta Trivedi · Accepted Answer

f: R+ â†’ [âˆ’5, âˆž) given by f(x) = 9x2 + 6x âˆ’ 5 to show that function is one-one we will show it by the contradiction let us take function is not one-one therefore there exist two or more numbers for which images are same we will take two different numbers $x_{1}, x_{2} \in R +$ let $f (x_{1}) = f (x_{2})$ $9 {x_{1}}^{2} + 6 x_{1} - 5 = 9 {x_{2}}^{2} + 6 x_{2} - 5 9 {x_{1}}^{2} + 6 x_{1} = 9 {x_{2}}^{2} + 6 x_{2} 9 {x_{1}}^{2} - 9 {x_{2}}^{2} + 6 x_{1} - 6 x_{2} = 0 9 ({x_{1}}^{2} - {x_{2}}^{2}) + 6 (x_{1} - x_{2}) = 0 (x_{1} - x_{2}) [9 (x_{1} + x_{2}) + 6] = 0$ since $x_{1} a n d x_{2}$ are positive therefore $9 (x_{1} + x_{2}) + 6 > 6$ it cannot be zero therefore $x_{1} - x_{2} = 0 \Rightarrow x_{1} = x_{2}$ therefore it contradicts our assumption hence the function is one-one A function f:Xâ†’Y is onto if for every yâˆˆY there exist a preimage in X let y= 9x2 + 6x âˆ’ 5 $9 x^{2} + 6 x - 5 - y = 0 w h e r e x \in R + x = \frac{- 6 + \sqrt{36 + 36 (5 + y)}}{2 * 9} = \frac{- 6 + 6 \sqrt{6 + y}}{18} x = \frac{- 1 + \sqrt{6 + y}}{3}$ $f (x) = 9 * (\frac{- 1 + \sqrt{6 + y}}{3})^{2} + 6 * (\frac{- 1 + \sqrt{6 + y}}{3}) - 5 = 9 * (\frac{1 + 6 + y - 2 \sqrt{6 + y}}{9}) + 2 (- 1 + \sqrt{6 + y}) - 5 = 7 + y - 2 \sqrt{6 + y} - 2 + 2 \sqrt{6 + y} - 5 = 7 + y - 7 = y$ hence f(x)=y therefore f(x) is onto

Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is one one,onto hence invertible

Consider f: R+ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is one one,onto hence invertible