Chapter 13, Problem 1P

Given the formula

$f\left(x\right)=-x^2+8x-12$

(a) Determine the maximum and the corresponding value of x for this function analytically (i.e., using differentiation).

(b) Verify that Eq. (13.7) yields the same results based on initial guesses of $x_0=0,x_1=2,$ and $x_2=6$.

To determine

To calculate: The maximum value of the function and the corresponding value of variable $x$ for the function,

$f\left(x\right)=-x^2+8x-12$

Answer to Problem 1P

Solution:

The maximum value of the function $f\left(x\right)=-x^2+8x-12$ is $4$ at $x=4$.

Explanation of Solution

Given Information:

The function $f\left(x\right)$ is given as,

$f\left(x\right)=-x^2+8x-12$

Calculation:

Evaluate the first order derivative of the function $f\left(x\right)$.

$f\text{'}\left(x\right)=-2x+8$ …… (1)

Equate equation (1) to zero and evaluate critical point.

$\begin{array}{c}-2x+8=0\\ x=\frac{-8}{-2}\\ =4\end{array}$

Therefore, a critical point is at $x=4$.

Evaluate the second order derivative of the function $f\left(x\right)$ to check if the function attains a maximum or minimum value at the critical point.

$f\text{'}\text{'}\left(x\right)=-2$ …… (2)

The sign of the second order derivative at $x=4$ is negative, therefore, at $x=4$ the given function attains a maximum value.

Substitute $x=4$ in the function $f\left(x\right)$ to find the maximum value of the function.

$\begin{array}{c}f\left(x\right)=-{\left(4\right)}^2+8\left(4\right)-12\\ =-16+32-12\\ =32-28\\ =4\end{array}$

Hence, the maximum value of the function at $x=4$ is $4$.

To determine

To prove: The results obtained in part (a) using the equation as follows:

$x_3=\frac{f\left(x_0\right)\left(x_1^2-x_2^2\right)+f\left(x_1\right)\left(x_2^2-x_0^2\right)+f\left(x_2\right)\left(x_0^2-x_1^2\right)}{2f\left(x_0\right)\left(x_1^{}-x_2^{}\right)+2f\left(x_1\right)\left(x_2^{}-x_0^{}\right)+2f\left(x_2\right)\left(x_0^{}-x_1^{}\right)}$

using the initial guess of $x_0=0,x_1=2\text{ and }{x}_2=6$.

Explanation of Solution

Given Information:

The function is given as,

$f\left(x\right)=-x^2+8x-12$

The initial guesses are $x_0=0,x_1=2\text{ and }{x}_2=6$.

Formula used:

The expression for the parabolic interpolation is given as,

$x_3=\frac{f\left(x_0\right)\left(x_1^2-x_2^2\right)+f\left(x_1\right)\left(x_2^2-x_0^2\right)+f\left(x_2\right)\left(x_0^2-x_1^2\right)}{2f\left(x_0\right)\left(x_1^{}-x_2^{}\right)+2f\left(x_1\right)\left(x_2^{}-x_0^{}\right)+2f\left(x_2\right)\left(x_0^{}-x_1^{}\right)}$

Here, $x_0,x_1\text{and }{x}_2$ are the initial guesses and $f\left(x_0\right),f\left(x_1\right)\text{and }f\left(x_2\right)$ are the values of the function at the respective initial guesses.

Proof:

Rewrite the equation of the parabolic interpolation:

$x_3=\frac{f\left(x_0\right)\left(x_1^2-x_2^2\right)+f\left(x_1\right)\left(x_2^2-x_0^2\right)+f\left(x_2\right)\left(x_0^2-x_1^2\right)}{2f\left(x_0\right)\left(x_1^{}-x_2^{}\right)+2f\left(x_1\right)\left(x_2^{}-x_0^{}\right)+2f\left(x_2\right)\left(x_0^{}-x_1^{}\right)}$ …… (3)

Substitute $x=0$ in $f\left(x\right)$ to get the value of $f\left(x_0\right)$.

$\begin{array}{c}f\left(x_0\right)=-{\left(0\right)}^2+8\left(0\right)-12\\ =-12\end{array}$

Thus, the value of $f\left(x_0\right)$ is $-12$.

Substitute $x=2$ in $f\left(x\right)$ to get the value of $f\left(x_1\right)$.

$\begin{array}{c}f\left(x_1\right)=-{\left(2\right)}^2+8\left(2\right)-12\\ =-4+16-12\\ =0\end{array}$

Thus, the value of $f\left(x_1\right)$ is $0$.

Substitute $x=6$ in $f\left(x\right)$ to get the value of $f\left(x_2\right)$.

$\begin{array}{c}f\left(x_2\right)=-{\left(6\right)}^2+8\left(6\right)-12\\ =-36+48-12\\ =0\end{array}$

Thus, the value of $f\left(x_2\right)$ is $0$.

Substitute $0$ for $x_0$, $2$ for $x_2$, $6$ for $x_2$, $-12$ for $f\left(x_0\right)$, $0$ for $f\left(x_1\right)$ and $0$ for $f\left(x_2\right)$ in equation (3).

$\begin{array}{c}{x}_3=\frac{\left(-12\right)\left[{\left(2\right)}^2-{\left(6\right)}^2\right]+\left(0\right)\left[{\left(6\right)}^2-{\left(0\right)}^2\right]+\left(0\right)\left[{\left(0\right)}^2-{\left(2\right)}^2\right]}{2\left(-12\right)\left[\left(2\right)-\left(6\right)\right]+2\left(0\right)\left[\left(6\right)-\left(0\right)\right]+2\left(0\right)\left[\left(0\right)-\left(2\right)\right]}\\ =4\end{array}$

Thus, the value of $x_3$ is $4$, this value matches with the results obtained in part (a).