# Finding range of $a$

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

• A non-continuous function can also have (local) extrema. Sep 24, 2018 at 10:34
• 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 Sep 24, 2018 at 10:36
• Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step. Sep 24, 2018 at 10:38
• I miswrote that when double checked with my textbook. Sorry for the inconvenience Sep 24, 2018 at 10:41

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

• Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience Sep 24, 2018 at 14:11

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 $$a=-1$$

• 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 Sep 24, 2018 at 11:08
• 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. Sep 24, 2018 at 11:27
• you solution is partially correct look at my solution hope you understand Sep 24, 2018 at 12:58