MHB Difference Quotient of f(x)=1/(x-3)

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The difference quotient for the function f(x) = 1/(x-3) is calculated using the formula (f(x+h) - f(x))/h. Substituting f(x) into the formula leads to f(x+h) = 1/((x+h)-3). After simplification, the difference quotient becomes (1/((x+h)-3) - 1/(x-3))/h. Common mistakes often arise during the simplification process, particularly in combining fractions. Correctly applying algebraic techniques is essential for obtaining the accurate difference quotient.
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Simply and find the difference quotient for f(x)= 1/(x-3)

I know the difference quotient formula is f(x+h)-f(x)/h but when I try solving it I keep getting it wrong.
 
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kendalgenevieve said:
Simply and find the difference quotient for f(x)= 1/(x-3)

I know the difference quotient formula is f(x+h)-f(x)/h but when I try solving it I keep getting it wrong.

Hi kendalgenevieve!

How is it wrong?
What did you get?
 

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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