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

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

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$h$ is increasing whenever $f$ is increasing

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(A) Given $h(x) = f(x) - (f(x))^2 + (f(x))^3$. Differentiating with respect to $x$,we get: $h'(x) = f'(x) - 2f(x)f'(x) + 3(f(x))^2 f'(x)$ Factoring out $f'(x)$: $h'(x) = f'(x) [1 - 2f(x) + 3(f(x))^2]$ To analyze the quadratic expression $3(f(x))^2 - 2f(x) + 1$,we complete the square: $3(f(x))^2 - 2f(x) + 1 = 3 \left( (f(x))^2 - \frac{2}{3}f(x) + \frac{1}{3} \right)$ $= 3 \left( (f(x) - \frac{1}{3})^2 - \frac{1}{9} + \frac{1}{3} \right)$ $= 3 \left( (f(x) - \frac{1}{3})^2 + \frac{2}{9} \right)$ Since $(f(x) - \frac{1}{3})^2 \ge 0$,the expression $(f(x) - \frac{1}{3})^2 + \frac{2}{9}$ is always positive. Therefore,$h'(x) = 3f'(x) \left( (f(x) - \frac{1}{3})^2 + \frac{2}{9} \right)$. Since the term in the bracket is always positive,the sign of $h'(x)$ is the same as the sign of $f'(x)$. Thus,$h$ is increasing whenever $f$ is increasing.

Let $h(x) = f(x) - (f(x))^2 + (f(x))^3$ for every real number $x$. Then

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