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Help finding the local min in a piecewise function

thinkbot

New member
Dec 14, 2013
5
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Dec 14, 2013
5
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
Jan 29, 2012
1,151
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 [tex]16- x^2[/tex] 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.