MHB Finding Slope of Tangent Line: How-To Guide

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To find the slope of a tangent line, it's essential to understand that only straight lines have a constant slope, while curves have varying slopes at different points. The discussion illustrates this by analyzing three straight line segments: from (-5, 5) to (-2, 2), from (-2, 2) to (0, 4), and from (0, 4) to (4, 2). Each segment has a distinct slope, which can be calculated using the formula (y2 - y1) / (x2 - x1). Understanding these concepts is crucial for grasping how to find slopes in both linear and non-linear contexts. Mastering this will enhance comprehension of calculus and its applications.
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My professor shown us in class how to find the slope but I still don't understand
 

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Find which slope? Surely your professor, when showing how to find slope explained that only straight lines have a single slope. Other curves, including "broken line curves" like this, have different slopes at different points. Here there are three different straight lines, one from (-5, 5) to (-2, 2), another from (-2, 2) to (0, 4), and the third from (0, 4) to (4, 2). What is the slope of each of those?
 

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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