Science:Math Exam Resources/Courses/MATH100/December 2019/Question 4 (b)

MATH100 December 2019

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[hide] Question 4 (b)
Suppose that we know that the second Maclaurin polynomial for f(x) is T₂(x)=2 + 3 x². What is the second Maclaurin polynomial for the function e^x f(x)?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
The Maclaurin polynomial is just another name for the Taylor polynomial centered at 0.

[show] Hint 2
The second order Maclaurin polynomial of a differentiable function f(x) is $T_{2}(x)=f(0)+f'(0)x+{\frac {f''(x)}{2!}}x^{2}$ .

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[show] Solution
The Maclaurin polynomial is the same as the Taylor polynomial centered at 0. From inspection of T₂(x)=2 + 3 x² we know that $f(0)=2,\,f'(0)=0\,{\text{ and }}f''(0)=2!\cdot 3.$ On the other hand, the second Maclaurin polynomial for g(x) = e^x f(x) has the form $T_{2,g}(x)=g(0)+g'(0)x+{\frac {g''(0)}{2!}}x^{2}$ . Here, since e⁰=1, we have: * g(0) = f(0)=2, * g'(x) = e^xf(x) + e^xf'(x), so that g'(0)= f(0) + f'(0)=2; * g"(x) = e^xf(x) + 2e^xf'(x) + e^xf"(x) . This results in g"(0) = f(0) + 2f'(0) + f"(0)=8. Putting everything together: $T_{2,g}(x)=2+2x+4x^{2}$ .