For the real values of x, maximum value of   xx2−5x+9  is

Let y=xx2−5x+9   ⇒yx2−(5y+1)x+9y=0 Here a=y,b=−(5y+1),c=9y Δ≥0⇒(5y+1)2−4.y.9y≥0       (∵b2−4ac≥0)                 ⇒11y2−10y−1≤0               ⇒(11y+1)(y−1)≤0               ⇒y∈[−111,1]                   y=1, −111 ∴  Maximum value =1