Chapter 4, Problem 49RE

To determine

Tofind:theabsolute maximum value of the function using Newtonian method and correctup-to 8 decimal places.

Answer to Problem 49RE

Theabsolute maximum value of the function is $\text{f}\left(0.33541803\right)\text{=1}\text{.16718557}\text{.}$

Explanation of Solution

Given:

  $\text{f}\left(\text{t}\right){\text{=cost+t-t}}^{\text{2}}$ .

Concept used:

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{>0}\text{.}$ the concave will be open upward and local minima can be found.

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{<0}\text{.}$ the concave will open downward and local maxima can be found.

Newton’s method formula:

  $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$ .

Calculation:

The function is similar to a downward opening parabola so the local max will be the absolute maximum. Here need to find the zero of $\text{<msup><mo>f</mo><mo>′</mo></msup>}$ .

  $\text{f}\left(\text{t}\right){\text{=cost+t-t}}^{\text{2}}$ .

  $\text{<msup><mo>f</mo><mo>′</mo></msup>}\left(\text{t}\right)\text{=-sint+1-2t}$

  $\text{lets consider the function interms of x}$ .

  $\text{g}\left(x\right)=-\sin x+1-2x$ .

  $\text{<msup><mo>g</mo><mo>′</mo></msup>}\left(x\right)=-\cos x-2$ .

The graph of $\text{f}\left(\text{t}\right){\text{=cost+t-t}}^{\text{2}}$

  

Newton’s method formula:

  $x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}$ .

  $x_1=0.3$ .

  $x_2=0.3-\frac{f\left(0.3\right)}{f^{\prime }\left(0.3\right)}=0.3353529264$ .

  $x_3=0.3353529264-\frac{f\left(0.3353529264\right)}{f^{\prime }\left(0.3353529264\right)}=0.3354180321$ .

  $x_4=0.3354180321-\frac{f\left(0.3354180321\right)}{f^{\prime }\left(0.3354180321\right)}=0.3354180324$ .

  $\text{<msup><mo>f</mo><mo>″</mo></msup>}\left(\text{t}\right)\text{=-cost-2}$ is negative for all $\text{t=0}\text{.33541803}$ .is where the absolute maximum is located.

The function value is

  $\text{f}\left(0.33541803\right)\text{=cos}\left(0.33541803\right)\text{+}0.33541803\text{-}{\left(0.33541803\right)}^{\text{2}}$ .

  $\Rightarrow \text{f}\left(0.33541803\right)\text{=1}\text{.16718557}\text{.}$

Hence the absolute maximum value of the function is $\text{f}\left(0.33541803\right)\text{=1}\text{.16718557}\text{.}$