MHB Help finding the local min in a piecewise function

Click For Summary
The discussion centers on finding the local minimum of the piecewise function defined as f(x) = {16 - x^2 for -4 <= x <= 0, 2x - 3 for 0 <= x <= 4. The first derivative test indicates a critical point at x=0, but there is confusion due to the function having two values at that point. It is noted that the derivative of the first piece is -2x, which equals zero at x=0, while the second piece's derivative is never zero. Consequently, potential local minima could occur at x = -4, x = 0, or x = 4, but the exact nature of f(0) complicates the determination of a local minimum.
thinkbot
Messages
5
Reaction score
0
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 don't think that's right. Any help?
 
Physics news on Phys.org
Re: Help finding the local min in a peicewise function

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

Do you find any critical values inside the two sub-domains?
 
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
 
Re: Help finding the local min in a peicewise function

thinkbot said:
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 don't think that's 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.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
View full post »

Similar threads

Graduate How to Find Critical Points of function f(x,y,z)
  • · Replies 9 ·
Replies
9
Views
3K
Undergrad Understanding a proof of inexistence of max nor min
  • · Replies 4 ·
Replies
4
Views
2K
MHB Is f(x)=ln(x)/x Increasing or Decreasing?
  • · Replies 1 ·
Replies
1
Views
2K
I have troubles finding the limit of this piecewise function
  • · Replies 9 ·
Replies
9
Views
2K
MHB How can I ensure continuity for a piecewise function with a radical term?
  • · Replies 2 ·
Replies
2
Views
2K
MHB 243.14.7.5 Find all local extreme values
  • · Replies 1 ·
Replies
1
Views
1K
MHB 141.30 how many points of inflection will the graph of the function have
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Uniform convergence of family of functions with continuous index
  • · Replies 3 ·
Replies
3
Views
2K
High School How do I invert this exponential function?
  • · Replies 3 ·
Replies
3
Views
4K
MHB Decide volume given two functions
  • · Replies 3 ·
Replies
3
Views
1K