Use the generalised mean value theorem to prove this

[chipotleaway]

Homework Statement

Let f(x) be a continuous function on [a, b] and differentiable on (a, b). Using the generalised mean value theorem, prove that:

f(x)=f(c) + (x-c)f'(c)+\frac{(x-c)^2}{2}f''(\theta) for some \theta \in (c, x)

Homework Equations

Hints given suggest consdiering F(x) = f(x) - f(c) - f'(c)(x-c) and G(x) = (x-a)^2
(F and G and their derivatives are the functions that appear in the given mean value theorem)

The Attempt at a Solution

I don't have no clue as to how to proceed other than making the substitutions in the hints and plugging it into the MVT equation and hoping the function I'm after pops out but this doesn't seem right - especially for the number of marks it's worth (plus I'm not really getting anywhere with it anyway).

 

[micromass]

What is the generalized mean value theorem?

 

[chipotleaway]

Cauchy's mean value theorem. If F and G are continuous on [a, b] and differentiable on (a, b) and G'(x)≠0 for all x in (a, b) then there exists some c in (a, b) such that:

\frac{F'(c)}{G'(c)}=\frac{F(b)-F(a)}{G(b)-G(a)}

 

[micromass]

Cauchy's mean value theorem. If F and G are continuous on [a, b] and differentiable on (a, b) and G'(x)≠0 for all x in (a, b) then there exists some c in (a, b) such that:

\frac{F'(c)}{G'(c)}=\frac{F(b)-F(a)}{G(b)-G(a)}

OK, then you should just plug in your functions $F(x)=f(x)-f(x)-f^\prime(c)(x-c)$ and $G(x) = (x-c)^2$ into this mean-value theorem.

Note however, that the numbers $c$, $a$ and $b$ in this mean-value theorem are not the same as the ones in the OP!

 

[chipotleaway]

Thanks! I'll have to check my algebra then, haha.

 

[chipotleaway]

If F(x)=f(x)-f(c)-f'(c)(x-c), then when I differentiate, do I treat f(c) and f'(c) as constants? I'm getting F'(x)=f'(x)-f''(c)(x-c)+f'(c)

 

[micromass]

If F(x)=f(x)-f(c)-f'(c)(x-c), then when I differentiate, do I treat f(c) and f'(c) as constants?

Yes. They are constants.