Contents

• 1  Prove that it has a unique local extreme in $\left[1,2\right]$
• 2  Approximate the minimum by Newton’s method. Use as initial guess $x_{0}=1$ and perform four iterations
• 3  Compute the residual

Consider the function $$h(x)=2x^{2}-x^{3}+\ln\left(2+x\right).$$

1. Prove that it has a unique local extreme in $\left[1,2\right].$
2. Approximate the minimum by Newton’s method. Use as initial guess $x_{0}=1$ and perform four iterations.
3. Compute the residual.

Prove that it has a unique local extreme in $\left[1,2\right]$

If we are looking for the extremes of $h,$ since the necessary condition for extreme for a differentiable function is that $h'(x)=0$, we are looking for the zeros of $h'.$

$$h'(x) = 4x-3x^2+\frac{1}{2+x}= \frac{(4x-3x^2)(2+x)+1}{2+x}= \frac{-3x^3-2x^2+8x+1}{2+x}$$

The zeros of $h'$ are the zeros of $f$

$$f(x) = -3x^3-2x^2+8x+1$$

$$f(x) = -3x^3-2x^2+8x+1$$

We see that the three sufficient conditions to exist a single root in the interval are met:

1. $f$ is continuous.
2. $f$ has different signs at the endpoints.
3. $f$ is strictly increasing (or decreasing) in this interval.

For our function:

1. $f$ is a polynomial and therefore, it is continuous.
2. $f(1)=4$ and $f(2)=-15$
3. $f$ is strictly decreasing in $[1,2]$ because $f'<0$ in $(1,2)$, as we can see if we factorize the derivative

$$ f'(x) = -9x^2-4x+8 $$

We compute the zeros of this second degree polynomial $ax^2+bx+c=0$ with

$$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

as $a=-9$, $b=-4$ and $c=8$ we have

$$x_{1,2} = \frac{4\pm \sqrt{(-4)^2-4(-9)(8)}}{2(-9)}\quad x_1= -1.19 \quad x_2 = 0.75$$

Therefore

$$f'(x)=a(x-x_1)(x-x_2)=-9(x+1.19)(x-0.75)$$

and, as the first factor, $-9$, is negative, and the second ,$(x+1.2)$, and third, $(x-0.75)$, positive in $(1,2)$

$$f'(x)= -9(x+1.2)(x-0.74) = (-)(+)(+)\lt 0 \quad \mathrm{in} \quad (1,2)$$

Approximate the minimum by Newton’s method. Use as initial guess $x_{0}=1$ and perform four iterations

The iterations are done following the formula

$$x_{k+1} = x_k-\frac{f(x_k)}{f'(x_k)}$$

In this case

$$f(x) = -3x^3-2x^2+8x+1 \quad f'(x) = -9x^2-4x+8 $$

That is

$$x_{k+1} =x_k-\frac{-3x_k^3-2x_k^2+8x_k+1}{-9x_k^2-4x_k+8}$$

Iterations

• $x_0=1$, $$x_1 =1-\frac{-3(1)^3-2(1)^2+8(1)+1}{-9(1)^2-4(1)+8}=1-\frac{-3-2+8+1}{-9-4+8}=1+\frac{4}{5}=1.8$$

• $x_1=1.8$, $$x_2 =1.8-\frac{-3(1.8)^3-2(1.8)^2+8(1.8)+1}{-9(1.8)^2-4(1.8)+8}=1.497602$$

And so on

\begin{array}{|rc|} \hline k & x_k \\ \hline 0 & 1.000000\\ 1 & 1.800000\\ 2 & 1.497602\\ 3 & 1.410600\\ 4 & 1.403193\\ \hline \end{array}

Once Loop Reflect

Compute the residual

The value of the function at the root is zero. The value of the function in the approximation is called residual and, if it is a good approximation, it is a small

$$r =|f(x_4)|=0.0008$$

Also, the residual in step $k$ is known, while the error is unknown (we need the exact solution, which is what we are looking for).

As $\alpha= 1.403140$ the absolute error is

$$e_a = |x_4-\alpha|=|1.403193-1.403140|=0.00005$$