I need to show that if $f$ is differentiable in $(a, b)$ and continuous in $[a, b]$ then there exists $c \in (a, b)$ such that: $$\frac{a \cdot f(a) - b \cdot f(b)}{a - b} = f(c) + c \cdot f'(c)$$ Ok so I know that there exists a point $c \in (a, b)$ so that $\frac{f(a) - f(b)}{a - b} = f'(c)$ but I don't understand how to get to the given equation... What is $f(c)$? and what is $c \cdot f'(c)$?
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2$\begingroup$ Try $g(x)=xf(x)$. $\endgroup$– AbstractionDec 30, 2015 at 12:31
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2 Answers
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Hint:
The right hand side looks like the derivative of
$$xf(x)$$
evaluated at $c$. How about using the theorem on this new function $xf(x)$?
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Assume $$g(x)=xf(x)$$
Now $g(x)$ is continuous and differentiable in the interval $[a,b]$
So, applying Lagrange's Mean Value Theorem for $$g(x)=xf(x)$$ we have that $$\frac{bf(b)-af(a)}{b-a}=\left[\frac{d}{dx}\{xf(x)\}\right]_{x=c}=f(c)+cf'(c)$$ for some $c \in (a,b)$.
This implies that there exists $c \in (a, b)$ such that: $$\frac{a \cdot f(a) - b \cdot f(b)}{a - b} = f(c) + c \cdot f'(c)$$
answered Dec 30, 2015 at 12:32
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$\begingroup$ Thanks! Why is $g(x)$ continuous and differentiable in the same intervals? $\endgroup$– LisaDec 30, 2015 at 13:27
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$\begingroup$ @Lisa You're welcome. $g(x)$ is continuous and differentiable in the same intervals as $x$ is continuous and differentiable in [a,b] and it is given that $f(x)$ is continuous and differentiable in [a,b]. And the product of 2 continuous and differentiable functions is always continuous and differentiable. $\endgroup$ Dec 30, 2015 at 13:31