Mathematics - Functions Question with Solution | TestHub

Given : $f(x)=2x^3-15x^2+36x+1$ $\Rightarrow f^{\prime }(x)=6x^2-30x+36$ $=6\left(x^2-5x+6\right)=6(x-2)(x-3)$ $\because f^{\prime }(x)>0\forall x\in [0,2)$ and $f^{\prime }(x)<0\forall x\in (2,3)$ $\therefore f(x)$ is increasing on $[0,2)$ and decreasing on $(2,3)$ $\therefore f(x)$ is many one on $[0,3]$ Also $f(0)=1,f(2)=29,f(3)=28$ $\therefore$ Absolute min $=1$ and Absolute max $=29$ $\therefore$ Range of $f=[1,29]=$ codomain Hence $f$ is onto.