给我一个解题视频---6. The function $f$ has derivatives of all orders for all real numbers. It is known that $f(0) = 2$, $f'(0) = 3$, $f''(x) = -f(x^2)$, and $f'''(x) = -2x \cdot f'(x^2)$.
(a) Find $f^{(4)}(x)$, the fourth derivative of $f$ with respect to $x$. Write the fourth-degree Taylor polynomial for $f$ about $x = 0$. Show the work that leads to your answer.
(b) The fourth-degree Taylor polynomial for $f$ about $x = 0$ is used to approximate $f(0.1)$. Given that $|f^{(5)}(x)| \leq 15$ for $0 \leq x \leq 0.5$, use the Lagrange error bound to show that this approximation is within $\frac{1}{10^5}$ of the exact value of $f(0.1)$.
(c) Let $g$ be the function such that $g(0) = 4$ and $g'(x) = e^x f(x)$. Write the second-degree Taylor polynomial for $g$ about $x = 0$.
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We have a function f with derivatives of all orders for all real numbers. We're given that f of 0 equals 2, f prime of 0 equals 3, the second derivative f double prime of x equals negative f of x squared, and the third derivative f triple prime of x equals negative 2x times f prime of x squared. Our goal is to find the fourth derivative and construct the fourth-degree Taylor polynomial about x equals 0.
To find the fourth derivative, we differentiate f triple prime of x equals negative 2x times f prime of x squared with respect to x. Using the product rule, we get f to the fourth of x equals negative 2 times f prime of x squared plus negative 2x times the derivative of f prime of x squared. By the chain rule, the derivative of f prime of x squared is f double prime of x squared times 2x. This gives us f to the fourth of x equals negative 2 f prime of x squared minus 4 x squared f double prime of x squared.
Now we evaluate each derivative at x equals 0. We already know f of 0 equals 2 and f prime of 0 equals 3. For f double prime of 0, we use the relation f double prime of x equals negative f of x squared, so f double prime of 0 equals negative f of 0 squared, which equals negative f of 0, giving us negative 2. For f triple prime of 0, we have negative 2 times 0 times f prime of 0 squared, which equals 0. Finally, f to the fourth of 0 equals negative 2 f prime of 0 minus 4 times 0 squared times f double prime of 0, which simplifies to negative 2 times 3, giving us negative 6.
Now we construct the fourth-degree Taylor polynomial using the formula P 4 of x equals f of 0 plus f prime of 0 times x plus f double prime of 0 over 2 factorial times x squared plus f triple prime of 0 over 3 factorial times x cubed plus f to the fourth of 0 over 4 factorial times x to the fourth. Substituting our values: 2 plus 3x plus negative 2 over 2 times x squared plus 0 over 6 times x cubed plus negative 6 over 24 times x to the fourth. Simplifying the fractions, we get 2 plus 3x minus x squared minus one fourth x to the fourth. This is our final fourth-degree Taylor polynomial for f about x equals 0.
Let's summarize our solution for part a. We found the fourth derivative by differentiating f triple prime of x using the product rule and chain rule, giving us f to the fourth of x equals negative 2 f prime of x squared minus 4 x squared f double prime of x squared. We then evaluated all derivatives at x equals 0 and constructed the fourth-degree Taylor polynomial: P 4 of x equals 2 plus 3x minus x squared minus one fourth x to the fourth. This completes part a of the problem, providing both the fourth derivative formula and the Taylor polynomial expansion about x equals 0.