Find the first and second degree Maclaurin polynomials for the function  f(x)=5ex+3e−x centered at  a=0 .Select the correct answer below:T1(x)=8+3x,T2(x)=8+3x+4x2 T1(x)=8+2x,T2(x)=8+2x+4x2 T1(x)=8+2x,T2(x)=8+2x−4x2 T1(x)=8−2x,T2(x)=8−2x+4x2

The Maclaurin series is a Taylor series expansion of a function about 0. The general form of a Taylor series is:

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ...

For a Maclaurin series, a = 0, so the series simplifies to:

f(x) = f(0) + f'(0)x + f''(0)x^2/2! + ...

Given the function f(x) = 5e^x + 3e^-x, we first need to find f(0), f'(0), and f''(0).

f(0) = 5e^0 + 3e^0 = 5 + 3 = 8

f'(x) = 5e^x - 3e^-x, so f'(0) = 5e^0 - 3e^0 = 5 - 3 = 2

f''(x) = 5e^x + 3e^-x, so f''(0) = 5e^0 + 3e^0 = 5 + 3 = 8

So, the first degree Maclaurin polynomial (T1) is:

T1(x) = f(0) + f'(0)x = 8 + 2x

And the second degree Maclaurin polynomial (T2) is:

T2(x) = f(0) + f'(0)x + f''(0)x^2/2! = 8 + 2x + 8x^2/2 = 8 + 2x + 4x^2

So, the correct answer is: T1(x) = 8 + 2x, T2(x) = 8 + 2x + 4x^2