1. **State the problem:** Find the degree 3 Taylor polynomial $T_3(x)$ of the function $$f(x) = (-7x + 109)^{\frac{5}{4}}$$ at the point $a=4$. 2. **Recall the Taylor polynomial formula:** $$ T_3(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 $$ We need to find $f(a)$, $f'(a)$, $f''(a)$, and $f'''(a)$. 3. **Calculate $f(a)$:** $$ f(4) = (-7 \cdot 4 + 109)^{\frac{5}{4}} = ( -28 + 109 )^{\frac{5}{4}} = 81^{\frac{5}{4}} $$ Since $81 = 3^4$, $$ 81^{\frac{5}{4}} = (3^4)^{\frac{5}{4}} = 3^{4 \cdot \frac{5}{4}} = 3^5 = 243 $$ 4. **Find the first derivative $f'(x)$:** Using the chain rule, $$ f'(x) = \frac{5}{4}(-7x + 109)^{\frac{5}{4} - 1} \cdot (-7) = -\frac{35}{4}(-7x + 109)^{\frac{1}{4}} $$ Evaluate at $x=4$: $$ f'(4) = -\frac{35}{4} (81)^{\frac{1}{4}} = -\frac{35}{4} \cdot 3 = -\frac{105}{4} $$ 5. **Find the second derivative $f''(x)$:** Differentiate $f'(x)$: $$ f''(x) = -\frac{35}{4} \cdot \frac{1}{4} (-7x + 109)^{\frac{1}{4} - 1} \cdot (-7) = -\frac{35}{4} \cdot \frac{1}{4} \cdot (-7) (-7x + 109)^{-\frac{3}{4}} $$ Simplify constants: $$ f''(x) = \frac{245}{16} (-7x + 109)^{-\frac{3}{4}} $$ Evaluate at $x=4$: $$ f''(4) = \frac{245}{16} (81)^{-\frac{3}{4}} = \frac{245}{16} \cdot 3^{-3} = \frac{245}{16} \cdot \frac{1}{27} = \frac{245}{432} $$ 6. **Find the third derivative $f'''(x)$:** Differentiate $f''(x)$: $$ f'''(x) = \frac{245}{16} \cdot (-\frac{3}{4}) (-7x + 109)^{-\frac{3}{4} - 1} \cdot (-7) = \frac{245}{16} \cdot (-\frac{3}{4}) \cdot (-7) (-7x + 109)^{-\frac{7}{4}} $$ Simplify constants: $$ f'''(x) = \frac{245}{16} \cdot \frac{21}{4} (-7x + 109)^{-\frac{7}{4}} = \frac{5145}{64} (-7x + 109)^{-\frac{7}{4}} $$ Evaluate at $x=4$: $$ f'''(4) = \frac{5145}{64} (81)^{-\frac{7}{4}} = \frac{5145}{64} \cdot 3^{-7} = \frac{5145}{64} \cdot \frac{1}{2187} = \frac{5145}{139968} $$ 7. **Write the Taylor polynomial:** $$ T_3(x) = 243 - \frac{105}{4}(x - 4) + \frac{245}{432} \cdot \frac{(x - 4)^2}{2} + \frac{5145}{139968} \cdot \frac{(x - 4)^3}{6} $$ Simplify factorial denominators: $$ T_3(x) = 243 - \frac{105}{4}(x - 4) + \frac{245}{864}(x - 4)^2 + \frac{5145}{839808}(x - 4)^3 $$