The Maclaurin series for f(x) = 3x² cos(5x) is 3x²/2.
The Maclaurin series for the function f(x) = 3x² cos(5x) can be found by expressing it as a power series. The general formula for a Maclaurin series is:
f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ...
To find the Maclaurin series for f(x), we need to calculate its derivatives at x = 0. Let's start by finding the first few derivatives:
f(x) = 3x² cos(5x)
f'(x) = 6x cos(5x) - 15x² sin(5x)
f''(x) = 6 cos(5x) - 30x sin(5x) - 30x sin(5x) - 75x² cos(5x)
f'''(x) = -60 sin(5x) - 60 sin(5x) - 150x cos(5x) - 150x cos(5x) + 150x² sin(5x)
Now, let's evaluate these derivatives at x = 0:
f(0) = 3(0)² cos(5(0)) = 3(0) = 0
f'(0) = 6(0) cos(5(0)) - 15(0)² sin(5(0)) = 0
f''(0) = 6 cos(5(0)) - 30(0) sin(5(0)) - 30(0) sin(5(0)) - 75(0)² cos(5(0)) = 6
f'''(0) = -60 sin(5(0)) - 60 sin(5(0)) - 150(0) cos(5(0)) - 150(0) cos(5(0)) + 150(0)² sin(5(0)) = 0
Now we can plug these values into the Maclaurin series formula:
f(x) = 0 + 0x + 6x²/2! + 0x³/3! + ...
Simplifying this gives us:
f(x) = 3x²/2
So the Maclaurin series for f(x) = 3x² cos(5x) is 3x²/2.