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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:

  1. f(x) = x^2;
  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).

  1. 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*}

  2. 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*}

2 comments

    • 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.

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