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?
Best Answer
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 $$-\infty<x<3$$