A differentiable function $f(x)$ is such that $f(-0.1) = 2$, $f(0) = 1$, and $f(0.1) = 2$.

Then, $\dfrac{d^2f}{dx^2}$ at $x = 0$ is ______ (answer in integer).

Question

The value of the integral
\[ \int_{0.2}^{1.4} (\sin x - \log_e x + e^x) \, dx \]
using Simpson's three-eighth rule, by taking interval size $ h = 0.2 $, is:

Answer 1

Step 1: Write Taylor expansions around $x=0$.
For small $h$, $f(h) = f(0) + h f'(0) + \dfrac{h^2}{2}f''(0) + O(h^3)$ and $f(-h) = f(0) - h f'(0) + \dfrac{h^2}{2}f''(0) + O(h^3)$.

Step 2: Add the two expansions.
Adding cancels the odd $f'(0)$ term: $f(h) + f(-h) = 2f(0) + h^2 f''(0) + O(h^4)$.
Rearranging gives $f''(0) \approx \dfrac{f(h) + f(-h) - 2f(0)}{h^2}$, the same second derivative estimate reached from a different route.

Step 3: Plug in the numbers.
With $h = 0.1$: $f''(0) \approx \dfrac{2 + 2 - 2(1)}{(0.1)^2} = \dfrac{2}{0.01} = 200$.

Final Answer:
The Taylor series route confirms the same value for the second derivative.
\[ \boxed{200} \]

A differentiable function \(f(x)\) is such that \(f(-0.1) = 2\), \(f(0) = 1\), and \(f(0.1) = 2\).

Then, \(\dfrac{d^2f}{dx^2}\) at \(x = 0\) is ______ (answer in integer).

Show Hint

Correct Answer: 200

Solution and Explanation

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A differentiable function \(f(x)\) is such that \(f(-0.1) = 2\), \(f(0) = 1\), and \(f(0.1) = 2\).Then, \(\dfrac{d^2f}{dx^2}\) at \(x = 0\) is ______ (answer in integer).