In homework I'm supposed to show that this function is monotonic, or if it's monotonic at all:
$$f(x)=x^{2}-2x-1$$
I did it like that but I'm not sure at all if it's the correct way to do, I need confirmation or my teacher is mad I did wrong please:
$$f'(x)=0$$
$$f'(x)=2x-2$$
$$0=2x-2|+2$$
$$2x=2|:2$$
$$x=1$$
Because $1>0$ the function is monotonic increasing?
Please say if it's correct I don't want teacher makes me bad in class and other laugh again... I need present tomorrow in class :(
$\endgroup$ 86 Answers
$\begingroup$The function isn't monotonic:
$$f(0)=-1,\\f(1)=-2,\\f(2)=-1.$$
$\endgroup$ $\begingroup$$$f(x)=x^2-2x-1=(x-1)^2-2.$$
We can shift the independent variable by $1$ without changing the monotonic character of the function.
$$g(x):=f(x+1)=x^2-2.$$
This function is even ($g(x)=g(-x)$), so it cannot be monotonic (if there is an increasing section, by symmetry there is a decreasing section).
Note that
$$0<a<b\implies a^2<b^2$$ because you may multiply the respective sides of two inequalities when the terms are positive. And
$$a<b<0\implies0<-b<-a\implies a^2>b^2.$$
So $x^2$ is decreasing to $x=0$, then increasing.
By shifting, $(x-1)^2$ is decreasing to $x=1$ then increasing, and so is $(x-1)^2-2$.
$\endgroup$ $\begingroup$For monotonicity (increasing at least) we need the derivative positive everywhere. This is not the case. You should test that when $x > 1$ (note the strict inequality here) the derivative is positive. When $x < 1$ the derivative is negative. So, $f$ is monotone on either $(1, \infty)$ or $(-\infty, 1)$.
You did it, just needed a little more. No worries, all shall be well.
$\endgroup$ 5 $\begingroup$The function is not monotonic in $\mathbb{R}$, it is a U-shaped function with minimum point at $x = 1$. It is monotonic in the intervals $[ - \infty, 1[$ and $]1 , +\infty]$.
Don't worry your teacher would be proud of you.
$\endgroup$ $\begingroup$This function can't be monotonic, for $\displaystyle \lim_{x\to \pm\infty} f(x)= +\infty$... You can say (studying the derivative like you did) that for $x=1$ you have a minimum, than the function will be monotonic decreasing in $(-\infty, 1]$ and increasing in $[1, +\infty)$.
$\endgroup$ 4 $\begingroup$Note you can solve this without calculus.
Let $\epsilon > 0$.
Then $$f(x + \epsilon) = (x + \epsilon)^2 -2(x + \epsilon) -1$$
$$= x^2 +2\epsilon x + \epsilon^2 - 2x - 2\epsilon -1 $$
thus $$f(x+\epsilon) - f(x) = (x^2 - x^2) +(-2x +2x)+(-1+1) + \epsilon^2-2\epsilon+2x\epsilon$$ $$=\epsilon(2x-2+\epsilon)$$
This difference is positive when $x>1$ and negative when $x<1$.
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