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(Solved): a. Let \( f(x)=\alpha x^{2}+\beta x+\gamma \) be a quadratic function with \( \alpha, \beta, \gamm ...



a. Let \( f(x)=\alpha x^{2}+\beta x+\gamma \) be a quadratic function with \( \alpha, \beta, \gamma \in \mathbb{R} \). Consid

a. Let \( f(x)=\alpha x^{2}+\beta x+\gamma \) be a quadratic function with \( \alpha, \beta, \gamma \in \mathbb{R} \). Consider the closed interval \( [a, b] \) which has midpoint \( c=\frac{a+b}{2} \). Prove that the slope of the secant line joining \( (a, f(a)) \) to \( (b, f(b)) \) is equal to the slope of the tangent line to \( f(x) \) at the point \( c \). b. What you did in part a. was prove that the midpoint \( c \) is the same as the point \( c \) specified in the Mean Value Theorem applied to \( f(x) \) on the interval \( [a, b] \). This is a special property of quadratic functions. It does not work in general. Find a counter example to demonstrate this. Your counterexample should specify both the function and the interval you are looking at.


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Given function ????? f(x)=?x2+?x+?differentiate f(x) f?(x)=2?x+? Slope of tangent at c=a+b2 ?????? f?(c)=2?c+?=2?a+b2+?=?(a+b)+?
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