Find the Taylor polynomials $p_{1}, \ldots, p_{5}$ centered at $a=0$ for $f(x)=7 e^{-2 x}$.

Step 1 ：We are given the function \(f(x) = 7e^{-2x}\) and we are asked to find the Taylor polynomials centered at \(a=0\).

Step 2 ：The Taylor series of a function about a point can be found using the formula: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]

Step 3 ：We first need to find the first few derivatives of \(f(x)\) at \(x=0\).

Step 4 ：The derivatives of \(f(x)\) at \(x=0\) are \[f'(0) = -14, f''(0) = 28, f'''(0) = -56, f''''(0) = 112, f'''''(0) = -224\]

Step 5 ：We substitute these values into the Taylor series formula to get the Taylor polynomials.

Step 6 ：The Taylor polynomials \(p_{1}, \ldots, p_{5}\) centered at \(a=0\) for \(f(x)=7 e^{-2 x}\) are: \[p_{1} = 7\] \[p_{2} = 7 - 14x\] \[p_{3} = 14x^{2} - 14x + 7\] \[p_{4} = -\frac{28}{3}x^{3} + 14x^{2} - 14x + 7\] \[p_{5} = \frac{14}{3}x^{4} - \frac{28}{3}x^{3} + 14x^{2} - 14x + 7\]

Step 7 ：\(\boxed{p_{1} = 7, p_{2} = 7 - 14x, p_{3} = 14x^{2} - 14x + 7, p_{4} = -\frac{28}{3}x^{3} + 14x^{2} - 14x + 7, p_{5} = \frac{14}{3}x^{4} - \frac{28}{3}x^{3} + 14x^{2} - 14x + 7}\)