Analysis: Proving a unique mean value property of quadratic functions

Question

a. Show $ quadratic polynomial $f(x)=ax^2+bx+c $, point c in Mean Value is midpoint.

For a: I know Mean Value Theorem states that (conditions being satisfied f(x) is continuous on $[g,h]$, f(x) is differentiable on $(g,h)$. $f'(c)=f(h)-f(g)/(h-g)$. So I take derivative of our function and get $2ax+b$. $f'(c)=2ac+b$. Thus we have $2ac+b=f(h)-f(g)/(h-g)$. Solving for c we get: $c=((f(h)-f(g)/(h-g)) -b)/2a$ Plugging in, we get: $((ah^2+bh+c)-(ag^2+bg+c)/(h-g)-b)/2a$. And,I'm not getting c to be the midpoint, so I think I'm doing something completely wrong.

Answer 1

$$f'(c)=b+2ac$$

$$\frac{f(h)-f(g)}{h-g}=\frac{-a g^2+a h^2-b g+b h}{h-g}$$

$$b+2ac=\frac{-a g^2+a h^2-b g+b h}{h-g}$$

$$2ac=\frac{-a g^2+a h^2-b g+b h}{h-g}-b$$

$$2ac =\frac{-a g^2+a h^2-b g+b h-bh+bg}{h-g}$$

$$c=\frac{a(h+g)(h-g)}{2a(h-g)}$$

$$c=\frac{h+g}{2}$$

Hope this helps

Answer 2

a) $f'(x) = 2ax + b$

MVT: There exists $c \in (g,h)$ such that: $f'(c) = \frac {f(g) - f(h)}{g-h}$

$\frac {f(g) - f(h)}{g-h}$ simplifies to:

$\frac {ag^2 + bg + c - ah^2 - bh - c}{g-h}\\ \frac {a(g^2 - h^2) + b(g - h)}{g-h}\\ \frac {a(g - h)(g+h) + b(g - h)}{g-h}\\ a(g+h)$

$f'(x) = 2ac + b = a(g+h) + b\\ c = \frac {g+h}{2}$

Answer 3

If $f(x)=ax^2+bx+c$ then $f'(x)=2ax+b$. It follows that $$f(y)-f(x)=a(y^2-x^2)+b(y-x)=\left(2a{x+y\over2}+b\right)(y-x)=f'\left({x+y\over2}\right)(y-x)\ .$$ This proves that quadratic polynomials indeed have the alleged property.

Now the converse: Assume that a function $f:\>{\mathbb R}\to{\mathbb R}$ satisfies $$f(t+h)-f(t-h)=2h f'(t)\tag{1}$$ for all $t$ and all $h$. Since $f$ is differentiable it follows that $f'$ is differentiable as well, hence $f''(t)$ exists for all $t$. If we differentiate $(1)$ with respect to $h$ for fixed $t$ we obtain $$f'(t+h)+f'(t-h)=2f'(t)\ ,$$ and differentiating agin with respect to $h$ gives $$f''(t+h)-f''(t-h)=0\qquad \forall t, \ \forall h\ .$$ The last identity is saying that $f''$ is constant. It follows that any $f$ satisfying $(1)$ has to be a quadratic polynomial.