The function f:[0,3] rightarrow [1,29],define | vedclass

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(B) Given function: $f(x) = 2x^3 - 15x^2 + 36x + 1$ for $x \in [0, 3]$.Step $1$: Check for one-one property.Find the derivative: $f'(x) = 6x^2 - 30x +

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onto but not one-one

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(B) Given function: $f(x) = 2x^3 - 15x^2 + 36x + 1$ for $x \in [0, 3]$.Step $1$: Check for one-one property.Find the derivative: $f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x-2)(x-3)$.Since $f'(x)$ changes sign at $x=2$ within the interval $[0, 3]$,the function is not monotonic. Specifically,$f(x)$ increases on $[0, 2]$,decreases on $[2, 3]$,and increases on $[3, \infty)$. Since it is not strictly monotonic on $[0, 3]$,it is not one-one (many-one).Step $2$: Check for onto property.Find the range of $f(x)$ on $[0, 3]$.$f(0) = 2(0)^3 - 15(0)^2 + 36(0) + 1 = 1$.$f(2) = 2(8) - 15(4) + 36(2) + 1 = 16 - 60 + 72 + 1 = 29$.$f(3) = 2(27) - 15(9) + 36(3) + 1 = 54 - 135 + 108 + 1 = 28$.Since the function is continuous on $[0, 3]$,the range is $[min(f(0), f(2), f(3)), max(f(0), f(2), f(3))] = [1, 29]$.Since the range $[1, 29]$ is equal to the codomain $[1, 29]$,the function is onto.Conclusion: The function is onto but not one-one.

The function $f:[0,3] \rightarrow [1,29]$,defined by $f(x)=2x^3-15x^2+36x+1$,is

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