Solve the problem.
-Use a graphical method to determine the approximate interval for which the second order Taylor polynomial for ln (1 + x) at x = 0 approximates ln (1 + x) with an absolute error of no more than 0.04.
A) $- 0.1928 \leq x \leq 0.7063$
B) $- 0.5525 \leq x \leq 0.2640$
C) $- 0.5525 \leq x \leq 0.5525$
D) $- 0.4310 \leq \mathrm { x } \leq 0.5525$

Question

Solve the problem.
-Use a graphical method to determine the approximate interval for which the second order Taylor polynomial for ln (1 + x) at x = 0 approximates ln (1 + x) with an absolute error of no more than 0.04. 
A) $- 0.1928 \leq x \leq 0.7063$
B) $- 0.5525 \leq x \leq 0.2640$
C) $- 0.5525 \leq x \leq 0.5525$
D) $- 0.4310 \leq \mathrm { x } \leq 0.5525$

Quizplus · Accepted Answer

The second order Taylor polynomial for ln (1 + x) at x = 0 is given by:

$$ P_2(x) = x - \frac{x^2}{2} $$

To find the interval where this polynomial approximates ln (1 + x) with an absolute error of no more than 0.04, we need to find the values of x where:

$$ \left|ln(1+x)-P_2(x)\right| \leq 0.04 $$

Substituting the expression for P2(x), we get:

$$ \left|\ln(1+x)-x+\frac{x^2}{2}\right| \leq 0.04 $$

Using a graphing calculator, we can graph the function y = ln(1+x) - x + (x^2)/2 and then use the trace function to find the x-values where the function is equal to ±0.04. Alternatively, we can use algebraic techniques such as the quadratic formula to solve for the roots of the inequality:

$$ \left(\ln(1+x)-x+\frac{x^2}{2}\right)^2 \leq 0.04^2 $$

This gives us:

$$ -0.4310 \leq x \leq 0.5525 $$

Therefore, the answer is D.

Solve the Problem −0.1928≤x≤0.7063- 0.1928 \leq x \leq 0.7063−0.1928≤x≤0.7063 B) −0.5525≤x≤0.2640- 0.5525 \leq x \leq 0.2640−0.5525≤x≤0.2640

Solve the Problem $- 0.1928 \leq x \leq 0.7063$ B) $- 0.5525 \leq x \leq 0.2640$