Let h(x) = f(x) - {f(x)}^2 + {f(x)}^3 for eve | vedclass

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(D) Given $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$.Differentiating with respect to $x$,we get:$h'(x) = f'(x) - 2f(x)f'(x) + 3\{f(x)\}^2f'(x)$Factoring

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Both $(A)$ and $(C)$

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(D) Given $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$.Differentiating with respect to $x$,we get:$h'(x) = f'(x) - 2f(x)f'(x) + 3\{f(x)\}^2f'(x)$Factoring out $f'(x)$:$h'(x) = f'(x) [1 - 2f(x) + 3\{f(x)\}^2]$Let $y = f(x)$. Then the expression becomes:$h'(x) = f'(x) (3y^2 - 2y + 1)$Consider the quadratic expression $Q(y) = 3y^2 - 2y + 1$. The discriminant $D = (-2)^2 - 4(3)(1) = 4 - 12 = -8$.Since $D  0$,the quadratic $3y^2 - 2y + 1$ is always positive for all real $y$.Thus,$h'(x) = f'(x) 	imes (	ext{a positive quantity})$.This implies that the sign of $h'(x)$ is the same as the sign of $f'(x)$.Therefore,$h(x)$ is increasing whenever $f(x)$ is increasing $(f'(x) > 0 \implies h'(x) > 0)$ and $h(x)$ is decreasing whenever $f(x)$ is decreasing $(f'(x) Hence,both $(A)$ and $(C)$ are correct.

Let $h(x) = f(x) - \{f(x)\}^2 + \{f(x)\}^3$ for every real number $x$,then

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