We need to solve two problems. a. Show that the function $f(x) = x^5 - 5x^4 + 5x^3 - 1$ has a maximum value when $x=1$ and $x=0$, and a minimum value when $x=3$. b. Solve the system of linear equations using Cramer's rule: $p + 2q - r = 9$ $2p - q + 3r = -2$ $3p + 2q + 3r = 9$

1. Problem Description

We need to solve two problems.

a. Show that the function $f(x) = x^5 - 5x^4 + 5x^3 - 1$ has a maximum value when $x=1$ and $x=0$, and a minimum value when $x=3$.

b. Solve the system of linear equations using Cramer's rule:

$p + 2q - r = 9$

$2p - q + 3r = -2$

$3p + 2q + 3r = 9$

2. Solution Steps

a. To find the maximum and minimum values of the function $f(x) = x^5 - 5x^4 + 5x^3 - 1$, we need to find its critical points by taking the first derivative and setting it equal to zero.

$f'(x) = 5x^4 - 20x^3 + 15x^2$

Set $f'(x) = 0$:

$5x^4 - 20x^3 + 15x^2 = 0$

$5x^2(x^2 - 4x + 3) = 0$

$5x^2(x-1)(x-3) = 0$

The critical points are $x=0, x=1,$ and $x=3$.

Now, we need to find the second derivative to determine whether these critical points are local maxima or minima.

$f''(x) = 20x^3 - 60x^2 + 30x$

Evaluate $f''(x)$ at the critical points:

$f''(0) = 20(0)^3 - 60(0)^2 + 30(0) = 0$. Since $f''(0) = 0$, the second derivative test is inconclusive at $x=0$. We check the first derivative $f'(x)$ around $x=0$. For $x<0$ close to $0$, $f'(x)>0$ and for $x>0$ close to $0$, $f'(x)>0$. Thus, $x=0$ is neither a local maximum nor a local minimum.

$f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10$. Since $f''(1) < 0$, $x=1$ is a local maximum.

$f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 20(27) - 60(9) + 90 = 540 - 540 + 90 = 90$. Since $f''(3) > 0$, $x=3$ is a local minimum.

The problem states that $x=0$ is a maximum value, this is incorrect according to the second derivative test. Also, $f(0)=-1, f(1) = 1-5+5-1=0, f(3) = 3^5 - 5(3^4) + 5(3^3)-1 = 243 - 5(81) + 5(27) - 1 = 243 - 405 + 135 - 1 = -28$.

Thus, $f(1) > f(0)$ so $x=1$ yields the larger value.

b. To solve the system of linear equations using Cramer's rule:

$p + 2q - r = 9$

$2p - q + 3r = -2$

$3p + 2q + 3r = 9$

We calculate the determinant of the coefficient matrix:

$D = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 2 & 3 \end{vmatrix} = 1((-1)(3) - (3)(2)) - 2((2)(3) - (3)(3)) + (-1)((2)(2) - (-1)(3)) = 1(-3-6) - 2(6-9) - 1(4+3) = -9 - 2(-3) - 7 = -9 + 6 - 7 = -10$

Now, we replace the first column with the constants to find $D_p$:

$D_p = \begin{vmatrix} 9 & 2 & -1 \\ -2 & -1 & 3 \\ 9 & 2 & 3 \end{vmatrix} = 9((-1)(3) - (3)(2)) - 2((-2)(3) - (3)(9)) + (-1)((-2)(2) - (-1)(9)) = 9(-3-6) - 2(-6-27) - 1(-4+9) = 9(-9) - 2(-33) - 5 = -81 + 66 - 5 = -20$

Next, we replace the second column with the constants to find $D_q$:

$D_q = \begin{vmatrix} 1 & 9 & -1 \\ 2 & -2 & 3 \\ 3 & 9 & 3 \end{vmatrix} = 1((-2)(3) - (3)(9)) - 9((2)(3) - (3)(3)) + (-1)((2)(9) - (-2)(3)) = 1(-6-27) - 9(6-9) - 1(18+6) = -33 - 9(-3) - 24 = -33 + 27 - 24 = -30$

Finally, we replace the third column with the constants to find $D_r$:

$D_r = \begin{vmatrix} 1 & 2 & 9 \\ 2 & -1 & -2 \\ 3 & 2 & 9 \end{vmatrix} = 1((-1)(9) - (-2)(2)) - 2((2)(9) - (-2)(3)) + 9((2)(2) - (-1)(3)) = 1(-9+4) - 2(18+6) + 9(4+3) = -5 - 2(24) + 9(7) = -5 - 48 + 63 = 10$

Now we use Cramer's rule to solve for $p, q,$ and $r$:

$p = \frac{D_p}{D} = \frac{-20}{-10} = 2$

$q = \frac{D_q}{D} = \frac{-30}{-10} = 3$

$r = \frac{D_r}{D} = \frac{10}{-10} = -1$

3. Final Answer

a. $f(x) = x^5 - 5x^4 + 5x^3 - 1$ has a local maximum at $x=1$ and a local minimum at $x=3$. The nature of the critical point at $x=0$ cannot be determined from the second derivative test.

b. $p = 2, q = 3, r = -1$