Let f'(x) 0 and g'(x)

Question

(A) Given $f'(x) > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function on $\mathbb{R}$.Given $g'(x) < 0$ for all $x \in \mathbb{R}$

Vedclass · Accepted Answer

$f[g(x)] > f[g(x + 1)]$ and $g[f(x)] > g[f(x + 1)]$

Vedclass · Answer

(A) Given $f'(x) > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function on $\mathbb{R}$.Given $g'(x) Since $f(x)$ is strictly increasing,for $x Since $g(x)$ is strictly decreasing,applying $g$ to the inequality $f(x) $g(f(x)) > g(f(x+1))$.Similarly,for $x-1 Applying $g$ to this inequality gives $g(f(x-1)) > g(f(x))$,which is equivalent to $g(f(x)) For $f(g(x))$,since $g(x)$ is strictly decreasing,for $x  g(x+1)$.Applying $f$ (which is strictly increasing) preserves the inequality sign:$f(g(x)) > f(g(x+1))$.Thus,the correct statement is $f[g(x)] > f[g(x + 1)]$ and $g[f(x)] > g[f(x + 1)]$ is not necessarily true,but checking the options,option $A$ is the intended answer based on the logic of composition.

Let $f'(x) > 0$ and $g'(x) < 0$ for all $x \in \mathbb{R}$. Then which of the following is true?

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