1. **State the problem:** We want to find the second Taylor polynomial of the function $f(x) = \sqrt[4]{x} = x^{\frac{1}{4}}$ centered at $x = 7$. 2. **Find the derivatives:** - $f(x) = x^{\frac{1}{4}}$ - $f'(x) = \frac{1}{4}x^{-\frac{3}{4}}$ - $f''(x) = -\frac{3}{16}x^{-\frac{7}{4}}$ 3. **Evaluate function and derivatives at $x=7$:** - $f(7) = 7^{\frac{1}{4}}$ - $f'(7) = \frac{1}{4} \times 7^{-\frac{3}{4}}$ - $f''(7) = -\frac{3}{16} \times 7^{-\frac{7}{4}}$ 4. **Write the second Taylor polynomial $P_2(x)$ centered at $7$:** $$ P_2(x) = f(7) + f'(7)(x - 7) + \frac{f''(7)}{2}(x - 7)^2 $$ 5. **Taylor’s theorem for error bound:** The remainder term $R_2(x)$ satisfies $$ |R_2(x)| \leq \frac{M}{3!} |x - 7|^3 $$ where $M$ is an upper bound on $|f^{(3)}(c)|$ for some $c$ between $7$ and $x$. 6. **Calculate the third derivative:** $$ f^{(3)}(x) = \frac{105}{64} x^{-\frac{11}{4}} $$ 7. **Find $M$ for error bound:** Since $x$ near $7$, the maximum of $|f^{(3)}(x)|$ on an interval around $7$ is at the smaller endpoint if $x > 7$ or larger endpoint if $x < 7$ because of the negative exponent. This completes the steps.