7
$\begingroup$

I had to determine where the following function is increasing and where it's decreasing. I can figure those out, but how do I write it down with correct notation and how could I prove it?

$$ f:\mathbb{R}\to\ \mathbb{R} \qquad x\mapsto(x-3)^4 $$

I know that I can calculate the extremum/extrema by taking the second derivative:

$$ f''(x)=((x-3)^4)''=(4(x-3)^3)'=12(x-3)^2 $$

and taking finding its root(s):

$$ 12(x-3)^2=0 \\ (x-3)^2=0 \\ x-3=0 \\ x=3 $$

I know of course, that $\mathit{f}$ has an extremum, more specifically a minimum at this point ($x=3$)

And I can see from it's graph and by substituting values that it decreases on $]-\infty,3[$ and increases on $]3,+\infty[$

But how do I write this down and prove it? I was thinking about using sequences to prove, maybe?

$\endgroup$
  • 1
    $\begingroup$ Out of interest, you could take what you "see from the graph" and see what you get by substituting in points $x=3 \pm \epsilon$, where $\epsilon >0, <0$ both sides of the root. You would then get $f(3 \pm \epsilon) = (\pm \epsilon)^4 = \epsilon^4$. Its an equivalent way effectively by transitioning the function. Then $f'(3 \pm \epsilon)=4\epsilon^3$ which is strictly $> 0$ for $\epsilon>0$ and $<0$ similarly $\endgroup$ – MKF Sep 21 '16 at 16:06
  • 1
    $\begingroup$ You need to take a step back and really understand what the 1st derivative represents; it's gives you the slope of the tangent in a certain point x, as a function of x. If you plug in a certain x into the 1st derivative and it returns 0, it means the tangent is parallel to the x axis at this point. Likewise, positive value for a certain range of x values will mean the function is increasing. You got the result by accident, since what you had to do is find the roots of the 1st derivative, not 2nd. $\endgroup$ – Groo Sep 21 '16 at 20:32
  • $\begingroup$ @Groo I realise that now, I was confused because before doing this problem, I've done one involving finding inflection points, which require finding the root(s) of the second derivative. as an aside, in such a case, should I edit my original post? I am quite new to the site $\endgroup$ – bp99 Sep 21 '16 at 20:35
7
$\begingroup$

you have to solve the inequality $$f'(x)=4(x-3)^3\geq 0$$ or $$f'(x)=4(x-3)^3\le 0$$ thus your function is increasing if $$3\le x<+\infty$$ or decreasing if $$-\infty<x<3$$

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ You have given an unstrict inequality for the first half and a strict one for the second half, seemingly for no reason. $\endgroup$ – Micah Sep 21 '16 at 21:56
6
$\begingroup$

You find the extremum by taking the first derivative, not the second. You will still get $x=3$ in this case. Then to show the function is increasing on $]+3,\infty[$, you just need to show the first derivative is positive on the interval. So take the first derivative and show that.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

The big theorem connecting the derivative to monotonicity is:

If $f'$ is positive on an open interval, then $f$ is increasing on that interval (in fact, the closure of that interval). Likewise, if $f'$ is negative on an interval, then $f$ is decreasing on that interval.

In your case, the derivative is $f'(x) = 4(x-3)^3$. This is negative on $-\infty < x < 3$ and positive on $3 < x < \infty$. So $f$ is decreasing on $(-\infty,3]$ and increasing on $[3,\infty)$.

| cite | improve this answer | |
$\endgroup$

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.