Find the order 3 Taylor polynomial T3(x) of the given function given. Use exact values. 
f(x)=(-3x+87)^5/4 centered at a=2.

T3(x)= 


To find the order 3 Taylor polynomial \( T_3(x) \) of the function \( f(x) = \left(-3x + 87\right)^{\frac{5}{4}} \) centered at \( a = 2 \), we need to determine the function's value and the first three derivatives at \( x = 2 \). 

### Step-by-Step Solution:

#### 1. Evaluate \( f(2) \):

\[ f(2) = \left(-3(2) + 87\right)^{\frac{5}{4}} = \left(81\right)^{\frac{5}{4}} = (3^4)^{\frac{5}{4}} = 3^5 = 243 \]

#### 2. Compute the first derivative \( f'(x) \):

Using the chain rule:

\[ f(x) = \left(-3x + 87\right)^{\frac{5}{4}} \]

Let \( u = -3x + 87 \), then \( f(x) = u^{\frac{5}{4}} \).

\[ \frac{du}{dx} = -3 \]

\[ f'(x) = \frac{d}{dx} \left( u^{\frac{5}{4}} \right) = \frac{5}{4} u^{\frac{1}{4}} \cdot \frac{du}{dx} = \frac{5}{4} (-3x + 87)^{\frac{1}{4}} \cdot (-3) = -\frac{15}{4} (-3x + 87)^{\frac{1}{4}} \]

Evaluate \( f'(2) \):

\[ f'(2) = -\frac{15}{4} (81)^{\frac{1}{4}} = -\frac{15}{4} \cdot 3 = -\frac{45}{4} \]

#### 3. Compute the second derivative \( f''(x) \):

Using the chain rule and product rule:

\[ f'(x) = -\frac{15}{4} (-3x + 87)^{\frac{1}{4}} \]

\[ f''(x) = -\frac{15}{4} \cdot \frac{1}{4} (-3x + 87)^{-\frac{3}{4}} \cdot (-3) \]

\[ f''(x) = \frac{45}{16} (-3x + 87)^{-\frac{3}{4}} \cdot 3 \]

\[ f''(x) = \frac{135}{16} (-3x + 87)^{-\frac{3}{4}} \]

Evaluate \( f''(2) \):

\[ f''(2) = \frac{135}{16} \cdot (81)^{-\frac{3}{4}} \]

\[ (81)^{-\frac{3}{4}} = \left(3^4\right)^{-\frac{3}{4}} = 3^{-3} = \frac{1}{27} \]

\[ f''(2) = \frac{135}{16} \cdot \frac{1}{27} = \frac{135}{432} = \frac{15}{48} = \frac{5}{16} \]

#### 4. Compute the third derivative \( f'''(x) \):

\[ f''(x) = \frac{135}{16} (-3x + 87)^{-\frac{3}{4}} \]

\[ f'''(x) = \frac{135}{16} \cdot -\frac{3}{4} (-3x + 87)^{-\frac{7}{4}} \cdot (-3) \]

\[ f'''(x) = -\frac{405}{64} (-3x + 87)^{-\frac{7}{4}} \cdot 3 \]

\[ f'''(x) = -\frac{1215}{64} (-3x + 87)^{-\frac{7}{4}} \]

Evaluate \( f'''(2) \):

\[ f'''(2) = -\frac{1215}{64} \cdot (81)^{-\frac{7}{4}} \]

\[ (81)^{-\frac{7}{4}} = \left(3^4\right)^{-\frac{7}{4}} = 3^{-7} = \frac{1}{2187} \]

\[ f'''(2) = -\frac{1215}{64} \cdot \frac{1}{2187} = -\frac{1215}{139968} = -\frac{45}{5184} = -\frac{15}{1728} = -\frac{5}{576} \]

### Taylor Polynomial \( T_3(x) \):

\[ T_3(x) = f(2) + f'(2)(x-2) + \frac{f''(2)}{2!}(x-2)^2 + \frac{f'''(2)}{3!}(x-2)^3 \]

Substitute the values:

\[ T_3(x) = 243 - \frac{45}{4}(x-2) + \frac{5}{16 \cdot 2}(x-2)^2 - \frac{5}{576 \cdot 6}(x-2)^3 \]

\[ T_3(x) = 243 - \frac{45}{4}(x-2) + \frac{5}{32}(x-2)^2 - \frac{5}{3456}(x-2)^3 \]

\[ T_3(x) = 243 - \frac{45}{4}(x-2) + \frac{5}{32}(x-2)^2 - \frac{5}{3456}(x-2)^3 \]

This is the order 3 Taylor polynomial for the given function centered at \( a = 2 \).

Would you like further details on any step? Here are some related questions for practice:

1. How is the chain rule applied in differentiation?
2. What is the general form of a Taylor series expansion?
3. How can you find higher-order derivatives of complex functions?
4. What are the implications of the radius of convergence for a Taylor series?
5. How can you use Taylor polynomials to approximate functions near a given point?
6. What are the key differences between Taylor and Maclaurin series?
7. How do you determine the error bound for a Taylor polynomial approximation?
8. How can you apply Taylor series to solve real-world problems?

**Tip:** Always double-check your computations, especially when dealing with higher-order derivatives, to avoid errors in the polynomial expansion.

Learn how to find the order 3 Taylor polynomial T3(x) of the function f(x) = (-3x + 87)^{5/4} centered at a = 2. Detailed step-by-step solution including the evaluation of function and its derivatives. Suitable for advanced undergraduate level.

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