Question

The values of a function \(f(x)\) at three different values of \(x\) are given below:
\(f(0.90) = 7.75\), \(f(1.00) = 9.00\), \(f(1.10) = 10.35\).
Using the central difference approximation method, which is the CORRECT estimate of the first differential \(f'(1.00)\) and second differential \(f''(1.00)\) respectively?

• A. 10.00 and 13.00
• B. 13.00 and 10.00
• C. 10.00 and 10.00
• D. 13.00 and 13.00

Hint:

Use f'(x0) = (f(x0+h)-f(x0-h))/(2h) and f''(x0) = (f(x0+h)-2f(x0)+f(x0-h))/h^2.

Answer 1

The Correct Option is B

Step 1: Set up the problem using equal spacing $h = 0.10$:
The three given points $x = 0.90, 1.00, 1.10$ are equally spaced with $h = 0.10$, and the function values are $y_{-1}=7.75$, $y_0=9.00$, $y_{+1}=10.35$.
Step 2: Derive the first derivative using the symmetric slope idea:
The average rate of change from $x=0.90$ to $x=1.10$ estimates the slope at the midpoint $x=1.00$:$$\text{slope} = \frac{y_{+1} - y_{-1}}{2h} = \frac{10.35 - 7.75}{0.20} = \frac{2.60}{0.20} = 13.00$$So $f'(1.00) \approx 13.00$.
Step 3: Derive the second derivative using the curvature (concavity) idea:
The second derivative measures how the forward and backward slopes differ. The forward slope is:$$s_f = \frac{y_{+1}-y_0}{h} = \frac{10.35-9.00}{0.10} = \frac{1.35}{0.10} = 13.5$$The backward slope is:$$s_b = \frac{y_0-y_{-1}}{h} = \frac{9.00-7.75}{0.10} = \frac{1.25}{0.10} = 12.5$$
Step 4: Take the rate of change of slope over the interval $h$:
$$f''(1.00) \approx \frac{s_f - s_b}{h} = \frac{13.5 - 12.5}{0.10} = \frac{1.0}{0.10} = 10.00$$
Step 5: State the pair of results:
This confirms $f'(1.00) = 13.00$ and $f''(1.00) = 10.00$, matching option (B).
Final Answer:
\[ \boxed{f'(1.00) = 13.00, \ f''(1.00) = 10.00} \]