The function $$f__$$ satisfying $$f(b)$$−$$f(a)b$$−a≠f′(x) for any x∈(a$$,$$b) is

Question

The function $$f__$$ satisfying $$f(b)$$−$$f(a)b$$−a≠f′(x) for any x∈(a$$,$$b) is

Infinity Learn · Accepted Answer

The functions in (a) , (c) (d) satisfying hypothesis of  Lagrange’s Mean Value theorem so there is  c  ∈  (a$$, $$b)  such  that $$f(b)$$−$$f(a)b$$−$$a=f$$′(c). The function in (b) is  differentiable on (1$$,$$_2) but not continuous at $$x=1$$ and $$x=$$ $$2$$ as limx→$$1+$$ $$f(x)=limx$$→$$1+$$ $$x2=1$$≠$$f(1)$$ and limx→$$2$$− $$f(x)=limx$$→$$2$$− $$x2=4$$≠$$f(2)$$. Moreover, $$f(2)$$−$$f(1)2$$−$$1=$$−$$1$$ and f′(x)=2x$$>$$0$$ On $$(1$$,$$2) so there is no x∈(1$$,$$2) such that $$f(2)$$−$$f(1)2$$−$$1=f$$′(x)

$The function \underline{\underline{f}} satisfying \frac{f (b) - f (a)}{b - a} \neq f^{'} (x) for any x \in (a, b) is$

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The function f__ satisfying f(b)−f(a)b−a≠f′(x) for any x∈(a,b) is

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$The function \underline{\underline{f}} satisfying \frac{f (b) - f (a)}{b - a} \neq f^{'} (x) for any x \in (a, b) is$