3
$\begingroup$

If $$f(x) = \begin{cases} |x-2|+a^2-9a-9, &\text{if }x<2\\ 2x-3, &\text{if } x\geqslant2 \end{cases}$$ has local minima at $x=2$, then range of $a$ is… ?

My failed Attempt:

Wrote $$f(x) = \begin{cases} -x+a^2-9a-7, &\text{if }x<2\\ 2x-3, &\text{if } x\geqslant2 \end{cases}$$

The problem here is that I don't know whether the function is continious is required or not for finding maxima and minima. Not that I could check it if it was

Then My next step would be differentiating

$$f(x) = \begin{cases} -1, &\text{if }x<2\\ 2, &\text{if } x\geqslant2 \end{cases}$$

That means the function is decreasing before $x=2$ and then increasing. This is the best I could think for this question but it's not providing me any hint on how to proceed

$\endgroup$
4
  • $\begingroup$ A non-continuous function can also have (local) extrema. $\endgroup$
    – Karlo
    Sep 24, 2018 at 10:34
  • $\begingroup$ I always get confused at non continious functions. I haven't seen examples of them so that's why I am always little confused when dealing with them $\endgroup$
    – Scáthach
    Sep 24, 2018 at 10:36
  • 2
    $\begingroup$ Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step. $\endgroup$
    – Florian
    Sep 24, 2018 at 10:38
  • $\begingroup$ I miswrote that when double checked with my textbook. Sorry for the inconvenience $\endgroup$
    – Scáthach
    Sep 24, 2018 at 10:41

2 Answers 2

7
$\begingroup$

A non-continuous function can also have the minimum at $x=2$

Here you want a local minimum at $x=2$ thus $$ |2-2|+a^2-9a-9\ge 2\cdot2-3 $$

$$ a^2-9a-9\ge 1 $$ $$(a+1)(a-10)\ge 0$$

You get $$a\in(-\infty,-1]\cup[10,\infty)$$

You can look here on desmos for the simulation

$\endgroup$
1
  • $\begingroup$ Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience $\endgroup$
    – Scáthach
    Sep 24, 2018 at 14:11
2
$\begingroup$

There's no other way of doing this question except assuming it is continuous, otherwise why would they have given you the function broken like this at x=2?

So, put right hand limit and you get 4-3=1.
Put x=2 for x<2 to get $a^2$-9a-9.

Since the function is continuous, both limits will be same, i.e. 1
So, $$a^2-9a-9=1$$ $$ a^2-9a-10=0 $$ $$(a-10)(a+1)=0 $$ $$a=10$$ or $$a=-1$$

$\endgroup$
3
  • $\begingroup$ If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range $\endgroup$
    – Scáthach
    Sep 24, 2018 at 11:08
  • $\begingroup$ a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous. $\endgroup$ Sep 24, 2018 at 11:27
  • $\begingroup$ you solution is partially correct look at my solution hope you understand $\endgroup$ Sep 24, 2018 at 12:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .