Assume that $f: [0, 1] \to \mathbb{R}$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$ such that $f(x + h) = f(x) + hf'(x + \theta h)$ for some $0 < \theta < 1$. If $f(x) = x^2(1 + x)$, and $\theta$ is expressed in terms of $x$ and $h$, then the value of $$\lim_{h \to 0} \theta(x, h)$$ is
Mean Value Theorem Limit Calculation
The problem involves the Mean Value Theorem (MVT) stated as $f(x + h) = f(x) + hf'(x + \theta h)$, where $0 < \theta < 1$. The goal is to determine $\lim_{h \to 0} \theta(x, h)$ for the function $f(x) = x^2(1 + x)$.
Rearranging the MVT equation yields: $hf'(x + \theta h) = f(x+h) - f(x)$.
Taylor Series for MVT
We utilize the Taylor expansion of $f(x+h)$ around the point $x$: $f(x+h) = f(x) + hf'(x) + \frac{h^2}{2!}f''(x) + O(h^3)$
From this expansion, we have: $f(x+h) - f(x) = hf'(x) + \frac{h^2}{2}f''(x) + O(h^3)$.
Substitute this result into the rearranged MVT equation:
$hf'(x + \theta h) = hf'(x) + \frac{h^2}{2}f''(x) + O(h^3)$
Divide the entire equation by $h$ (assuming $h \neq 0$): $f'(x + \theta h) = f'(x) + \frac{h}{2}f''(x) + O(h^2) \quad (1)$
Next, apply the Taylor expansion to the term $f'(x + \theta h)$ around $x$, treating $f'$ as the function: $f'(x + \theta h) = f'(x) + (\theta h)f''(x) + O(h^2) \quad (2)$
Theta Value Determination
Equating the expressions for $f'(x + \theta h)$ from equations (1) and (2):
$f'(x) + \frac{h}{2}f''(x) + O(h^2) = f'(x) + \theta h f''(x) + O(h^2)$
By comparing the coefficients associated with the $h f''(x)$ term on both sides (and assuming $f''(x) \neq 0$), we get:
$\frac{1}{2} = \theta$
Limit of Theta Calculation
Let's verify this for the specific function $f(x) = x^2(1 + x) = x^2 + x^3$.
- The first derivative is $f'(x) = 2x + 3x^2$.
- The second derivative is $f''(x) = 2 + 6x$.
The condition $f''(x) \neq 0$ holds unless $x = -\frac{1}{3}$.
Since the derived value $\theta = \frac{1}{2}$ is a constant, it is independent of both $x$ and $h$. Therefore, the limit as $h$ approaches 0 is simply the constant value itself:
$ \lim_{h \to 0} \theta(x, h) = \lim_{h \to 0} \frac{1}{2} = \frac{1}{2} $
Important Questions from Mean Value Theorem
A series expansion for the function sin θ is
If f is the derivative of some function on [a, b], then there exists a number c in (a, b) such that Integral of f with respect to x =
Which condition is not required in checking for Taylor's theorem?
- What is the interval of Taylor series expansion of tan(x)?
According to the Mean Value Theorem, for a continuous function f(x) in the interval [a, b], there exists a value ξ in this interval such that \(\mathop \smallint \limits_a^b f\left( x \right)dx =\)
