THE FUNCTION f:[0,3] =>[1,29] DEFINED BY f(x)=2x^3-15x^2+36x+1,is

• ONE -ONE AND ONTO
• ONTO BUT 1-1
• 1-1 BUT NOT ONTO
• MANY-1

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}:\left[0,3\right]\to \left[1,29\right] \mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^3-15{\mathrm{x}}^2+36\mathrm{x}+1 \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=6{\mathrm{x}}^2-30\mathrm{x}+36=6\left({\mathrm{x}}^2-5\mathrm{x}+6\right) \\ \Rightarrow \mathrm{f}\text{'}\left(\mathrm{x}\right)=6\left(\mathrm{x}-2\right)\left(\mathrm{x}-3\right) \\ \mathrm{Now} \mathrm{we} \mathrm{observe} \mathrm{sign} \mathrm{change} \mathrm{of} \mathrm{f}\text{'}\left(\mathrm{x}\right) \mathrm{when} \mathrm{x}\in \left[0,3\right] \mathrm{using} \mathrm{method} \mathrm{of} \mathrm{interval} \\ 0++++\left(2\right)----\left(3\right) \\ \mathrm{Hence} \mathrm{function} \mathrm{increases} \mathrm{when} \mathrm{x}\in \left(0,2\right) \mathrm{and} \mathrm{then} \mathrm{decreases} \mathrm{when} \mathrm{x}\in \left(2,3\right) \\ \mathrm{Hence} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \mathrm{non}-\mathrm{monotonic} \mathrm{function}. \mathrm{Hence} \mathrm{it} \mathrm{is} \mathrm{not} \mathrm{one}-\mathrm{one}. \\ \mathrm{Now} \\ \mathrm{f}\left(0\right)=0+0+0+1=1 \\ \mathrm{f}\left(2\right)=2\times 2^3-15\times 2^2+36\times 2+1=29 \\ \mathrm{f}\left(3\right)=2\times 3^3-15\times 3^2+36\times 3+1=28 \\ \mathrm{Hence} \mathrm{range} \mathrm{is} \left[1,29\right] \mathrm{which} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{codomain}. \\ \mathrm{Hence} \mathrm{function} \mathrm{is} \mathrm{onto}. \end{gathered}$$

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