\( f(x)-[0,3] \rightarrow [1,29] \) then \( f(x)=2 x^{3}-15 x^{2}+36 x+1 \) is
Step-by-step Solution:
Here is the given mathematical expression formatted in MathJax LaTeX: \[ f'(x) = 6x^2 - 30x + 36 \] \[ = 6(x^2 - 5x + 6) \] \[ = 6(x - 2)(x - 3) \] \( f(x) \) is **increasing** in \( [0,2] \) and **decreasing** in \( [2,3] \). \( f(x) \) is **many-one**. \[ f(0) = 1, \quad f(2) = 29, \quad f(3) = 28 \] Range is: \[ [1,29] \] Hence, \( f(x) \) is **many-one onto**.