# How do I prove that a function decreases/increases on an interval?

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?

• 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 – MKF Sep 21 '16 at 16:06
• 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. – Groo Sep 21 '16 at 20:32
• @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 – bp99 Sep 21 '16 at 20:35

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 $$-\infty<x<3$$
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.
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 $[3,\infty)$.