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?
$\endgroup$ 33 Answers
$\begingroup$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 $$-\infty<x<3$$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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 $[3,\infty)$.
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