The function f:[0,3] arrow[1,29] , defined by f(x)=2 x^ 3 -15 x^ 2...

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