MHB What is the slope of functions at the edge case of $x=1$?

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The discussion focuses on determining the slope of functions at the edge case of $x=1$. Two functions are analyzed: $f(x)=x^2-1$ and $f(x)=2x+1$. For $f(x)=x^2-1$, the slope at $x=1$ is calculated, and it is confirmed that the graph includes the point $(1,0)$. Similarly, for $f(x)=2x+1$, the slope at $x=1$ is also evaluated, with a check on whether the graph includes the point $(1,0)$. Understanding these slopes at the edge case is crucial for analyzing function behavior.
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How can you solve something like this ?

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Sithira said:
How can you solve something like this ?

Hi Sithira! Welcome to MHB! ;)

We have 2 edge cases here around the point(s) where $x=1$.

Suppose we had $f(x)=x^2-1$.
What would the slope be at $x=1$?
Does its graph include the point $(1,0)$?

As for the second edge case, suppose we had $f(x)=2x+1$.
What would the slope be at $x=1$?
Does its graph include the point $(1,0)$?
 

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)}...
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