# Help finding the local min in a piecewise function

#### thinkbot

##### New member
f(x) = {
{ 16 - x^2, if -4 <= x <= 0
{ 2x - 3, if 0 <= x <= 4
I did the first derivative test and got zero but i dont think thats right. Any help?

#### MarkFL

Staff member
Re: Help finding the local min in a peicewise function

One quibble: $$\displaystyle f(0)$$ has two values defined.

Do you find any critical values inside the two sub-domains?

#### thinkbot

##### New member
Re: Help finding the local min in a peicewise function

f^1(x)={2x
{2
0=2x x=0
thats what i had but i think its wrong

#### HallsofIvy

##### Well-known member
MHB Math Helper
Re: Help finding the local min in a peicewise function

f(x) = {
{ 16 - x^2, if -4 <= x <= 0
{ 2x - 3, if 0 <= x <= 4
I did the first derivative test and got zero but i dont think thats right. Any help?
As MarkFL pointed out, you can have f(0)= 16 or -3 but not both.

The derivative of $$16- x^2$$ is -2x which is 0 at x= 0 which is a "break point". Of course, the derivative of 2x- 3 is never 0.

That tells you that a local minimum must occur at x= -4, x= 0, or x= 4. Without knowing if f(0) is 16 or -3, we cannot say which or even if f has a local minimum.