Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=e^{3 x}, \quad n=4 $$

Short answer

The Maclaurin polynomial of degree 4 for the function \( f(x) = e^{3x} \) is \( 1 + 3x + 4.5x^2 + 4.5x^3 + 3.375x^4 \)

Step-by-step solution

Calculate the Derivatives

The function given is \( f(x) = e^{3x} \). First, the derivative of the function must be found for each degree up to 4. \n1. for \( n=0 \), \( f^0(x) = e^{3x} \), thus \( f^0(0) = e^{3*0} = 1\)\n2. for \( n=1 \), \( f^1(x) = 3* e^{3x} \), thus \( f^1(0) = 3*e^{3*0} = 3 \)\n3. for \( n=2 \), \( f^2(x) = 9* e^{3x} \), thus \( f^2(0) = 9*e^{3*0} = 9 \)\n4. for \( n=3 \), \( f^3(x) = 27* e^{3x} \), thus \( f^3(0) = 27*e^{3*0} = 27 \)\n5. for \( n=4 \), \( f^4(x) = 81* e^{3x} \), thus \( f^4(0) = 81*e^{3*0} = 81 \)

Calculate Maclaurin Polynomial

All the needed derivatives have been calculated at zero. The next step will be to calculate the Maclaurin Polynomials using the formula mentioned above. \(\n P_4(x) = f^0(0) + f^1(0) * x/1! + f^2(0) * x^2/2! + f^3(0) * x^3/3! + f^4(0) * x^4/4! \).\nSubstituting the calculated derivatives at zero from step one: \( P_4(x) = 1 + 3x + 9x^2/2 + 27x^3/6 + 81x^4/24 \).\nSimplifying it leads to the final 4th degree Maclaurin Polynomial for the given function.