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[tex]\large\boxed{p(x) = 7x + 7x^2 + \frac{7}{2}x^3 + \frac{7}{6}x^4}[/tex]
Recall a Taylor series centered at x = 0:
[tex]p(x) = f(0) + f'(0)(x) + \frac{f''(0)}{2}x^{2} + \frac{f'''(0)}{3!}x^{3} + ...+ \frac{f^n}{n!}x^n[/tex]
Begin by finding the derivatives and evaluate at x = 0:
f(0) = 7(0)e⁰ = 0
f'(x) = 7eˣ + 7xeˣ f'(0) = 7e⁰ + 7(0)e⁰ = 7
f''(x) = 7eˣ + 7eˣ + 7xeˣ f''(0) = 7(1) + 7(1) + 0 = 14
f'''(x) = 7eˣ + 7eˣ + 7eˣ + 7xeˣ f'''(0) = 21
f⁴(x) = 7eˣ + 7eˣ + 7eˣ + 7eˣ + 7xeˣ f⁴(0) = 28
Now that we calculated 4 non-zero terms, we can write the Taylor series:
[tex]p(x) = 0 + 7x + \frac{14}{2}x^2 + \frac{21}{3!}x^3 + \frac{28}{4!}x^4[/tex]
Simplify:
[tex]p(x) = 7x + 7x^2 + \frac{7}{2}x^3 + \frac{7}{6}x^4[/tex]