Prove an alternate expression for the mean-value formula

Prove that the expression

$f(x+h) = f(x) + hf'(x+ \theta h) \qquad \text{where } 0 < \theta < 1$

is an equivalent form of the mean-value theorem.
Find the value of $\theta$ in terms of $x$ and $h$ when:

$f(x) = x^2$ ;
$f(x) = x^3$ .

For parts (a) and (b) keep $x$ fixed with $x \neq 0$ and find the limit of $\theta$ as $h$ tends to 0.

Proof. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then by the mean-value theorem we have

$f(b) - f(a) = f'(c) (b-a) \qquad \text{for some } c \in [a,b].$

Letting $a = x$ and $b = x + h$ for some $h > 0$ (since $b > a$ ), we have

$c \in [a,b] \quad \implies \quad c = x + \theta h \qquad \text{for some } \theta \in (0,1).$

(This follows since from our definitions, $h$ is the distance from $b-a$ . Then, since $c$ is somewhere in the interval $[a,b]$ its value must be $a$ plus some portion of the distance to $b$ . This portion is then $\theta$ , which is how we know $0 < \theta < 1$ .) Substituting $x = a$ and $x + h = b$ and $c = x + \theta h$ ,

$\begin{align*} f(b) - f(a) = f'(c) (b-a) && \implies && f(x+h) - f(x) &= f'(x+ \theta h)(x+h-x) \\ && \implies && f(x+h) &= f(x) + h f'(x + \theta h), \end{align*}$

where $0 < \theta < 1$ and $h > 0. \qquad \blacksquare$

Now for parts (a) and (b).

If $f(x) = x^2$ , we have $f'(x) = 2x$ , so,
$\begin{align*} f(x+h) = f(x) + hf'(x+ \theta h) && \implies && (x+h)^2 &= x^2 + 2h (x + \theta h) \\ && \implies && h^2 &= 2 \theta h^2 \\ && \implies && \theta &= \frac{1}{2} \\ && \implies && \lim_{h \to 0} \theta = \frac{1}{2}. \end{align*}$
If $f(x) = x^3$ , we have $f'(x) = 3x^2$ . So,
$\begin{align*} &&f(x+h) &= f(x) + hf'(x+ \theta h) \\ \implies && (x^3 + 3x^2 h + 3xh^2 + h^3) &= x^3 + 3hx^2 + 6 \theta x h^2 + 3 \theta^2 h^3. \\ \implies && 0 &= (3h^3)\theta^2 + (6xh^2) \theta - (3xh^2 + h^3) \\ \implies && 0 &= h \theta^2 + 2x \theta + \left( -x - \frac{h^2}{3} \right) \\ \intertext{Using the quadratic formula to solve for $\theta$ and noting that since $0 < \theta < 1$ only the positive root is possible, we continue computing...} \implies && \theta &= \frac{-2x + \sqrt{4x^2 + 4hx + 4h^2/3}}{2h} \\ \implies && \theta &= \frac{\sqrt{x^2 + xh + h^2/3} - x}{h} \\ \implies && \theta &= \frac{ \left( \sqrt{x^2 + xh + h^2/3} - x \right) \left( \sqrt{x^2 + xh + h^2/3} + x \right)}{h \left( \sqrt{x^2 + hx + h^2/3} + x \right)} \\ \implies && \theta &= \frac{x^2 + xh + \frac{h^2}{3} - x^2}{h \left( \sqrt{x^2 + hx + h^2/3} + x \right)} \\ \implies && \theta &= \frac{x + \frac{h}{3}}{x + \sqrt{x^2 + xh + h^2/3}} \\ \implies && \lim_{h \to 0} \theta &= \frac{1}{2}. \end{align*}$

5 comments

July 24, 2019 at 3:59 pm

M.d.S says:

on letter b
when you divide everything by h²
how come does the last term equals to -x-h²/3?
why is h still squared since it was to the third power and h³/h² = h¹

- February 19, 2021 at 5:50 am
  
  Mihajlo says:
  
  It is a typo, and it should be -x-h/3.
  
March 11, 2016 at 3:30 am

zz says:

Small nit: in part b, your solutions breaks when $x 5$ . When $x < 0$ , your approach using

$\theta = \frac{-2x - \sqrt{4x^2 + 4hx + 4h^2/3}}{2h}$

should work, since

$\lim_{h \to 0} \sqrt{x^2 + xh + h^2/3} = |x|.$

- March 11, 2016 at 5:26 am
  
  RoRi says:
  
  Hey, I think there’s a typo in the first line. The solution breaks when… $x=5$ ? I fixed up the other latex problem (sorry, there’s no way to edit your comment once it’s posted, and no way to preview the latex… I’m working on it). Let me know if I introduced errors.
  
  - July 4, 2019 at 4:29 pm
    
    Artem says:
    
    You have a typo in the quadratic formula. The discriminant should be $D = b^2 – 4ac$. So it should be $4h^3/3$ under the square root.

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Prove an alternate expression for the mean-value formula

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