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Math  /  Calculus

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Write a degree 3 Taylor Polynomial for f(x)=−4exf(x)=-4 e^{x} centered at x=−3x=-3. Let f(x)=−4exf(x)=-4 e^{x}. Find the function value and first 3 derivatives of f(x)f(x) at x=−3x=-3 f(x)=f(x)= □\square f(−3)=f(-3)= □\square f′(x)=f^{\prime}(x)= □\square f′(−3)=f^{\prime}(-3)= □\square f′′(x)=f^{\prime \prime}(x)= □\square f′′(−3)=f^{\prime \prime}(-3)= □\square f′′′(x)=f^{\prime \prime \prime}(x)= □\square f′′′(−3)=f^{\prime \prime \prime}(-3)= □\square thy

Studdy Solution

STEP 1

1. We are given the function f(x)=−4ex f(x) = -4 e^{x} .
2. We need to find the Taylor polynomial of degree 3 centered at x=−3 x = -3 .
3. We need to find the function value and the first three derivatives of f(x) f(x) at x=−3 x = -3 .

STEP 2

1. Evaluate the function f(x) f(x) at x=−3 x = -3 .
2. Find the first derivative of f(x) f(x) and evaluate it at x=−3 x = -3 .
3. Find the second derivative of f(x) f(x) and evaluate it at x=−3 x = -3 .
4. Find the third derivative of f(x) f(x) and evaluate it at x=−3 x = -3 .
5. Construct the Taylor polynomial of degree 3 using the values obtained.

STEP 3

Evaluate the function at x=−3 x = -3 :
f(−3)=−4e−3 f(-3) = -4 e^{-3}

STEP 4

Find the first derivative of f(x) f(x) :
f′(x)=ddx(−4ex)=−4ex f'(x) = \frac{d}{dx}(-4 e^{x}) = -4 e^{x}
Evaluate the first derivative at x=−3 x = -3 :
f′(−3)=−4e−3 f'(-3) = -4 e^{-3}

STEP 5

Find the second derivative of f(x) f(x) :
f′′(x)=ddx(−4ex)=−4ex f''(x) = \frac{d}{dx}(-4 e^{x}) = -4 e^{x}
Evaluate the second derivative at x=−3 x = -3 :
f′′(−3)=−4e−3 f''(-3) = -4 e^{-3}

STEP 6

Find the third derivative of f(x) f(x) :
f′′′(x)=ddx(−4ex)=−4ex f'''(x) = \frac{d}{dx}(-4 e^{x}) = -4 e^{x}
Evaluate the third derivative at x=−3 x = -3 :
f′′′(−3)=−4e−3 f'''(-3) = -4 e^{-3}

STEP 7

Construct the Taylor polynomial of degree 3 centered at x=−3 x = -3 :
The Taylor polynomial is given by:
P3(x)=f(−3)+f′(−3)(x+3)+f′′(−3)2!(x+3)2+f′′′(−3)3!(x+3)3P_3(x) = f(-3) + f'(-3)(x + 3) + \frac{f''(-3)}{2!}(x + 3)^2 + \frac{f'''(-3)}{3!}(x + 3)^3
Substitute the values obtained:
P3(x)=−4e−3−4e−3(x+3)+−4e−32(x+3)2+−4e−36(x+3)3P_3(x) = -4 e^{-3} - 4 e^{-3}(x + 3) + \frac{-4 e^{-3}}{2}(x + 3)^2 + \frac{-4 e^{-3}}{6}(x + 3)^3
The degree 3 Taylor polynomial for f(x)=−4ex f(x) = -4 e^{x} centered at x=−3 x = -3 is:
P3(x)=−4e−3−4e−3(x+3)−2e−3(x+3)2−23e−3(x+3)3P_3(x) = -4 e^{-3} - 4 e^{-3}(x + 3) - 2 e^{-3}(x + 3)^2 - \frac{2}{3} e^{-3}(x + 3)^3

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