0
$\begingroup$

I would like to inquire whether there is a simple way to prove that a function is decreasing or not. For example how would I prove that the function

$$Y = (X^.5 - 1)/0.5$$

is decreasing? I am not sure whether the negativity of the 2nd derivative solves this and would really like to understand the intuition better.

$\endgroup$
1
$\begingroup$

Intuitively, let's it's easier to go directly from the definition. Note that $\frac{(x^{1/2}-1)}{1/2} = 2 x^{1/2} - 2$. Note that $-2$ is constant. Now, consider $x_2 > x_1$ for any $x_2, x_1$ and compare $2x_2^{1/2}$ to $2x_1^{1/2}$:

$$\frac{2x_2^{1/2}}{2x_1^{1/2}} = \left(\frac{x_2}{x_1}\right)^{1/2}.$$

Since $x_2 > x_1$, we have

$$\left(\frac{x_2}{x_1}\right)^{1/2} > 1.$$

So starting at any point in the domain ($x_1$), if we go any distance to the right (to $x_2$), the function gets bigger. Now, simply show that it gets bigger without bound (in other words, it is not increasing to a limit of zero).

$\endgroup$
1
$\begingroup$

I am posting this answer because you wanted to know the monotonicity in terms of the derivative, otherwise this answer does not make any sense as there is already a nice answer of your question.

So if a function is differentiable then the monotonicity can be described as follows.

Let $f:[a,b]\to \Bbb R$ be continuous function which is differentiable on $(a,b)$, then

(i) $f^\prime(x)>0,\, \forall x\in (a,b)\implies f$ is (strictly) increasing on $[a,b]$

(ii) $f^\prime(x)<0,\, \forall x\in (a,b)\implies f$ is (strictly) decreasing on $[a,b]$

(iii) $f^\prime(x)=0,\, \forall x\in (a,b)\implies f$ is constant on $[a,b]$.

For your problem $f(x)=\frac{1}{0.5}(x^{0.5}-1)=2(\sqrt{x}-1)$. Clearly $f$ is differentiable on $(0,\infty)$ and $f^\prime(x)=\frac{1}{\sqrt{x}},\,\forall x\in(0,\infty)$. Also $f^\prime(x)>0$ for all $x\in(0,\infty)$. Thus $f$ is strictly increasing on $[0,\infty)$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.