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Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $\displaystyle \lim_{x \to +\infty} f'(x)=1$, then

1. $f$ is bounded
2. $f$ is increasing
3. $f$ is unbounded
4. $f'$ is bounded.
asked in Calculus | 107 views

+1 vote

a) False. At $+\infty$, the function is open, hence, it can not be bounded.

b) False. Only if $f'(x) > 0$ for all $x$ can we say that $f$ is increasing. This is not mentioned. What we know is that $f'(x)$ is positive near $+\infty$. So it could be the case that $f$ is decreasing near some value of $x$.

d) False. The function $f$ can have infinite slope for some $x$.

Ans- C

edited

What happens when the value of $f'$ is infinity for some $x \in \mathbb{R}$? Doesn't it make $f$ not differentiable at that value of $x$?

To get the correct explanation, think about the fact that $f'$ doesn't have to be infinite/non-existent at any $x \in \mathbb{R}$ to be unbounded.

Thanks @Pragy Sir. I got your point.

Can we say that, Since the given graph could be increasing and decreasing both. So f'(x) could be positive and negative both so summation of f'(x) over x∈ R could be infinite. Hence f'(x) is unbounded.

Can we say that, Since the given graph could be increasing and decreasing both. So $f'(x)$ could be positive and negative both so summation of $f'(x)$ over $x\in R$ could be infinite. Hence $f'(x)$ is unbounded.

1 - You can't really sum over a continuous set like $\mathbb{R}$. You integrate over it (integrals are limits of summations).

2 - How does the area under $f'$ being infinite relate to $f'$ being unbounded !?

The correct explanation is: $f'$ can be unbounded without being infinite/undefined at any point (thus making $f$ differentiable everywhere), if $f'$ tends to an infinite (positive or negative) as it approaches $-\infty$. That is, $\displaystyle \left | \lim_{x\to -\infty} f'(x) \right | = \infty$.

This way, $f'$ exists, and is finite on every value of $x \in \mathbb{R}$, but is still unbounded.

+1 vote