Question

Find the linear approximating polynomial for the following function centered at the given point a.point a. b. Find the quadratic approximating polynomial for the following function centered at the given point a. c. Use the polynomials obtained in parts​ (a) and​ (b) to approximate the given quantity. f (x )equals sine xf(x)=sinx​, aequals=negative StartFraction pi Over 4 EndFraction− π 4​; approximate sine (negative 0.23 pi )

Answer 1

Answer:

$(a)L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\(b)Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\(c)L(-0.23\pi)=-0.6626\\Q(-0.23\pi)=-0.6613$

Step-by-step explanation:

Given the function:

$f(x)=sin x, a=-\frac{\pi}{4}$

(a)Linear approximating polynomial

$L(x)=f(a)+f'(a)(x-a)\\f(a)=sin(-\frac{\pi}{4})=-\frac{\sqrt{2} }{2}\\f'(x)=cos x, a=-\frac{\pi}{4}\\f'(a)=cos (-\frac{\pi}{4})=\frac{\sqrt{2} }{2} \\Therefore:\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x-(-\frac{\pi}{4}))\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})$

(b)Quadratic approximating polynomial

$Q(x)=L(x)+\frac{1}{2}f''(a)(x-a)^2\\f''(x)=-sin(x), \\f''(a)=-sin(-\frac{\pi}{4})=\frac{\sqrt{2} }{2}\\Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2$

(c)When $x=-0.23\pi$

Using Linear Approximation polynomial

$L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\L(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})\\L(-0.23\pi)=-0.6626$

Using the Quadratic approximating polynomial

$Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\Q(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(-0.23\pi+\frac{\pi}{4})^2=-0.6613$