Let $P(x)=7-3(x-4)+5(x-4)^{2}-2(x-4)^{3}+6(x-4)^{4}$ be the fourth-degree Taylor polynomial for the function $f$ about 4 . Assume $f$ has derivatives of all orders for all real numbers.
(a) Find $f(4)$ and $f^{\prime \prime}(4)$.
(b) Write the second-degree Taylor polynomial for $f^{\prime}$ about 4 and use it to approximate $f^{\prime}(4.3)$.
(c) Write the fourth-degree Taylor polynomial for $g(x)=\int_{4}^{x} f(t) d t$ about 4 .
(d) Can $f(3)$ be determined from the information given? Justify your answer.

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $f(4)$, we evaluate the polynomial $P(x)$ at $x = 4$. ‖ ‖ step 2 ⋮ $P(4) = 7 - 3(4-4) + 5(4-4)^2 - 2(4-4)^3 + 6(4-4)^4 = 7$ ‖ ‖ step 3 ⋮ Therefore, $f(4) = 7$. ‖ ‖ step 4 ⋮ To find $f''(4)$, we need to differentiate $P(x)$ twice and evaluate at $x = 4$. ‖ ‖ step 5 ⋮ $P(x) = 7 - 3(x-4) + 5(x-4)^2 - 2(x-4)^3 + 6(x-4)^4$ ‖ ‖ step 6 ⋮ $P'(x) = -3 + 10(x-4) - 6(x-4)^2 + 24(x-4)^3$ ‖ ‖ step 7 ⋮ $P''(x) = 10 - 12(x-4) + 72(x-4)^2$ ‖ ‖ step 8 ⋮ $P''(4) = 10 - 12(4-4) + 72(4-4)^2 = 10$ ‖ ‖ step 9 ⋮ Therefore, $f''(4) = 10$. ‖ ‖ step 10 ⋮ To write the second-degree Taylor polynomial for $f'(x)$ about $x = 4$, we use $P'(x)$. ‖ ‖ step 11 ⋮ $P'(x) = -3 + 10(x-4) - 6(x-4)^2 + 24(x-4)^3$ ‖ ‖ step 12 ⋮ The second-degree Taylor polynomial for $f'(x)$ is $P'(x) \approx -3 + 10(x-4) - 6(x-4)^2$. ‖ ‖ step 13 ⋮ To approximate $f'(4.3)$, we evaluate the second-degree Taylor polynomial at $x = 4.3$. ‖ ‖ step 14 ⋮ $P'(4.3) \approx -3 + 10(4.3-4) - 6(4.3-4)^2$ ‖ ‖ step 15 ⋮ $P'(4.3) \approx -3 + 10(0.3) - 6(0.3)^2$ ‖ ‖ step 16 ⋮ $P'(4.3) \approx -3 + 3 - 0.54 = -0.54$ ‖ ‖ step 17 ⋮ Therefore, $f'(4.3) \approx -0.54$. ‖ ‖ step 18 ⋮ To write the fourth-degree Taylor polynomial for $g(x) = \int_{4}^{x} f(t) dt$ about $x = 4$, we integrate $P(x)$. ‖ ‖ step 19 ⋮ $g(x) = \int_{4}^{x} (7 - 3(t-4) + 5(t-4)^2 - 2(t-4)^3 + 6(t-4)^4) dt$ ‖ ‖ step 20 ⋮ $g(x) = 7(x-4) - \frac{3}{2}(x-4)^2 + \frac{5}{3}(x-4)^3 - \frac{1}{2}(x-4)^4 + \frac{6}{5}(x-4)^5$ ‖ ‖ step 21 ⋮ Therefore, the fourth-degree Taylor polynomial for $g(x)$ is $7(x-4) - \frac{3}{2}(x-4)^2 + \frac{5}{3}(x-4)^3 - \frac{1}{2}(x-4)^4 + \frac{6}{5}(x-4)^5$. ‖ ‖ step 22 ⋮ To determine if $f(3)$ can be found from the given information, we note that $P(x)$ is centered at $x = 4$. ‖ ‖ step 23 ⋮ Since $P(x)$ is a Taylor polynomial centered at $x = 4$, it does not provide information about $f(x)$ at $x = 3$. ‖ ‖ step 24 ⋮ Therefore, $f(3)$ cannot be determined from the given information. ‖ ∻Answer∻ ⚹ $f(4) = 7$, $f''(4) = 10$, $f'(4.3) \approx -0.54$, fourth-degree Taylor polynomial for $g(x)$ is $7(x-4) - \frac{3}{2}(x-4)^2 + \frac{5}{3}(x-4)^3 - \frac{1}{2}(x-4)^4 + \frac{6}{5}(x-4)^5$, and $f(3)$ cannot be determined. ⚹ ∻Key Concept∻ ⚹Taylor Polynomial Approximation⚹ ∻Explanation∻ ⚹Taylor polynomials provide approximations of functions around a specific point, but they do not give information about the function at points far from the center of the polynomial.⚹◊What is the formula for the second-degree Taylor polynomial for $f^{ \prime}$ about 4?⍭ Generate me a similar question◊

Sia