[Solved] Consider the MATLAB code:f = @(x) x^2; x = 2; h = 0.01; df = (f(x+h) - f(x-h)) / (2*h);What is the value of df approximately? | MATLAB

MATLAB - Numerical Methods

Consider the MATLAB code:

f = @(x) x^2;
x = 2;
h = 0.01;
df = (f(x+h) - f(x-h)) / (2*h);

What is the value of df approximately?

A5.00

B2.00

C4.00

D3.99

Step-by-Step Solution

Solution:

Step 1: Calculate f(x+h) and f(x-h)
f(2+0.01) = (2.01)^2 = 4.0401; f(2-0.01) = (1.99)^2 = 3.9601.
Step 2: Compute central difference
df = (4.0401 - 3.9601) / (2*0.01) = 0.08 / 0.02 = 4.00.
Step 3: MATLAB precision
In MATLAB, df is computed and displayed as 4.0000.
Final Answer:
4.00 -> Option C
Quick Check:
Central difference ≈ 4 [OK]

Quick Trick: Central difference approximates slope near x [OK]

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