Let phi (x) = (f(x))^3 - 3(f(x))^2 + 4f(x) + | vedclass

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(A) Differentiate $\phi(x)$ with respect to $x$:$\phi'(x) = 3(f(x))^2 f'(x) - 6f(x) f'(x) + 4f'(x) + 5 + 3 \cos x - 4 \sin x$Factor out $f'(x)$:$\phi'

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$\phi$ is increasing whenever $f$ is increasing

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(A) Differentiate $\phi(x)$ with respect to $x$:$\phi'(x) = 3(f(x))^2 f'(x) - 6f(x) f'(x) + 4f'(x) + 5 + 3 \cos x - 4 \sin x$Factor out $f'(x)$:$\phi'(x) = f'(x) [3(f(x))^2 - 6f(x) + 4] + 5 + 3 \cos x - 4 \sin x$Analyze the quadratic term $3(f(x))^2 - 6f(x) + 4$:The discriminant $D = (-6)^2 - 4(3)(4) = 36 - 48 = -12 Since the leading coefficient is positive,$3(f(x))^2 - 6f(x) + 4 > 0$ for all $f(x) \in R$.Analyze the trigonometric term $3 \cos x - 4 \sin x$:The range of $a \cos x + b \sin x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Thus,the range of $3 \cos x - 4 \sin x$ is $[-\sqrt{3^2 + (-4)^2}, \sqrt{3^2 + (-4)^2}] = [-5, 5]$.Therefore,$\phi'(x) = f'(x) \cdot [	ext{positive value}] + 5 + [	ext{value in } [-5, 5]]$.Since $5 + [-5, 5] \ge 0$,if $f'(x) \ge 0$ (i.e.,$f$ is increasing),then $\phi'(x) > 0$,meaning $\phi$ is increasing.

Let $\phi (x) = (f(x))^3 - 3(f(x))^2 + 4f(x) + 5x + 3 \sin x + 4 \cos x$ for all $x \in R$,then -

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