0
$\begingroup$

The question and its answer is given in the following 2 pictures:

enter image description here

enter image description here

The answer for me is not illustrative, I want a justification why should I delete each wrong choice, could anyone help me?

$\endgroup$
1
$\begingroup$

Quite simply, the other choices don't make any sense. Let's go through them one by one:

(B): The hypotheses are indeed satisfied. We need:

  1. f to be continuous on the closed interval
  2. f' to exist on the open interval

f' exists on the open interval as hypothesis. Since f' exists on $(0,\infty)$, it must exist on the closed subset $[1,2]$. That in turns means f is differentiable on $[1,2]$, which implies it is continuous. Ditto for $[2,x]$

(C): f' is strictly increasing, which means that:

$$a>b \Leftrightarrow f'(a) > f'(b)$$

Hence (3) is completely valid.

(D): $(x-2) > 0$, by choice of x. We can easily go from (3) to (4) with some manipulations. (4) is valid.

(E): (4) does imply (5). This is a standard result in analysis. The inequality holds for all $x>2$.

EDIT: For (E), you may use the following definition of the divergence of a function at infinity to prove that $$(\forall x, f(x) \ge g(x) \land\lim_{x \to \infty} g(x) = \infty) \Rightarrow \lim_{x \to \infty}f(x) = \infty $$

Let $ f: \mathbb{R} \rightarrow\mathbb{R}$ $$\lim_{x \to \infty}f(x) = \infty \Leftrightarrow \forall M>0, \exists c ,\forall x > c, f(x) > M$$

Source: Infinite limits

$\endgroup$
2
  • $\begingroup$ why you said " (4) does imply (5). This is a standard result in analysis."? $\endgroup$
    – user426277
    Jul 3 '17 at 13:14
  • 1
    $\begingroup$ @Idonotknow I have added a definition that will allow you to see the results for yourself. $\endgroup$ Jul 3 '17 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy