Math  /  Calculus

QuestionFind the Taylor polynomial T3(x)T_{3}(x) for the function ff centered at the number aa. f(x)=ex,a=1T3(x)=\begin{array}{l} f(x)=e^{x}, \quad a=1 \\ T_{3}(x)=\square \end{array}
Graph ff and T3T_{3} in the same viewing rectangle.

Studdy Solution

STEP 1

1. We are finding the third-degree Taylor polynomial for the function f(x)=ex f(x) = e^x .
2. The polynomial is centered at a=1 a = 1 .
3. We will use the Taylor series expansion formula to find the polynomial.

STEP 2

1. Recall the formula for the Taylor series expansion.
2. Calculate the necessary derivatives of f(x)=ex f(x) = e^x .
3. Evaluate these derivatives at x=a=1 x = a = 1 .
4. Construct the Taylor polynomial T3(x) T_3(x) .
5. Graph both f(x) f(x) and T3(x) T_3(x) .

STEP 3

The Taylor series expansion for a function f(x) f(x) about a point a a is given by: Tn(x)=f(a)+f(a)(xa)+f(a)2!(xa)2+f(a)3!(xa)3+ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots

STEP 4

Calculate the derivatives of f(x)=ex f(x) = e^x : - f(x)=ex f(x) = e^x - f(x)=ex f'(x) = e^x - f(x)=ex f''(x) = e^x - f(x)=ex f'''(x) = e^x

STEP 5

Evaluate these derivatives at x=1 x = 1 : - f(1)=e1=e f(1) = e^1 = e - f(1)=e1=e f'(1) = e^1 = e - f(1)=e1=e f''(1) = e^1 = e - f(1)=e1=e f'''(1) = e^1 = e

STEP 6

Substitute these values into the Taylor series formula: T3(x)=e+e(x1)+e2!(x1)2+e3!(x1)3 T_3(x) = e + e(x-1) + \frac{e}{2!}(x-1)^2 + \frac{e}{3!}(x-1)^3
Simplify the expression: T3(x)=e+e(x1)+e2(x1)2+e6(x1)3 T_3(x) = e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3

STEP 7

Graph f(x)=ex f(x) = e^x and T3(x)=e+e(x1)+e2(x1)2+e6(x1)3 T_3(x) = e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3 in the same viewing rectangle.
The third-degree Taylor polynomial T3(x) T_3(x) is: T3(x)=e+e(x1)+e2(x1)2+e6(x1)3 T_3(x) = e + e(x-1) + \frac{e}{2}(x-1)^2 + \frac{e}{6}(x-1)^3

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