MHB Finding a Min or Max point without knowing the function

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The discussion centers on analyzing the function h(x) = f(g(x)), where f is monotonically increasing and g has a local minimum at x=0. The derivative h'(x) is calculated using the chain rule, leading to h'(0) = 0 due to g'(0) = 0. It is concluded that h(x) has a local minimum at x=0, as h is locally increasing for x > 0 and locally decreasing for x < 0. While f does not influence the location of extrema, it does affect their values and the locations of zeroes. Thus, the relationship between f and g is crucial for understanding the behavior of h.
Yankel
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Hello,

I have this simple problem:

f(x) and g(x) are both differentiable. f is monotonically increasing for every x. g has a local min at x=0. we define h to be h(x)=f(g(x)).

Can we say anything about x=0 for h(x) ?

I used the chain rule to find that h'(x) = f'(g(x))*g'(x). at x=0 g'(x)=0, so h'(0) = 0 as well. is it possible to say if x=0 is min, max, or isn't is possible ?

thank you in advance.
 
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We can tell that g(x) is locally increasing for x > 0. Therefore h(x) is locally increasing for x > 0 as well.
And g(x) is locally decreasing for x < 0, so h(x) is locally decreasing for x < 0 as well.
Consequently h(x) has a local minimum at x=0.
Differentiability of any of the functions is not needed.
 
right, so f has no effect on h ?
 
Yankel said:
right, so f has no effect on h ?

Well, f doesn't affect the locations of the extrema, but it does affect the value of those extrema.
And f will also affect the locations of zeroes.
 
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)}...

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