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CatWhisperer
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Homework Statement
Let ##f:[a,b] \rightarrow R## be a differentiable function. Show that if ##P = \{ x_0 , x_1 , ... , x_n \}## is a partition of ##[a,b]## then $$L(P,f')=\sum_{j=1}^n m_j \Delta x_j \leq f(b) - f(a)$$ where ##m_j=inf \{ f'(t) : t \in [x_{j-1} , x_j ] \}## and ##\Delta x_j = x_j - x_{j-1}## for each ##1 \leq j \leq n##. Hint: Use the Mean-value Theorem.
Homework Equations
Mean-value Theorem:
If ##f## is continuous on ##[a,b]## and differentiable on ##(a,b)## then there exists a ##c \in (a,b)## such that $$f'(c) = \frac{f(b)-f(a)}{b-a}$$
The Attempt at a Solution
##f## is differentiable on ##[a,b]##, so it must also be continuous on ##[a,b]##, so by the mean value theorem there exists a ##c \in [a,b]## such that $$f'(c) = \frac{f(b)-f(a)}{b-a}$$ $$\sum_{j=1}^n \Delta x_j = b-a$$ $$\Leftrightarrow \sum_{j=1}^n m_j \Delta x_j = (b-a)\sum_{j=1}^n m_j \leq f(b) - f(a)$$ $$\Leftrightarrow \sum_{j=1}^n m_j \leq \frac{f(b) - f(a)}{b-a} = f'(c)$$ (where ##f'(c)## is described by the MVT)
which is about as far as I have gotten. I would really appreciate any assistance. I don't really know how I am supposed to bring MVT into this; how I have done so here doesn't really seem to help me.
Thanks very much in advance.