(a) use newtons method to find the critical numbers of the function ( f(x)=x^{5}-x^{4}+2x^{3}-3x ) correct…

(a) use newtons method to find the critical numbers of the function ( f(x)=x^{5}-x^{4}+2x^{3}-3x ) correct to six decimal places. (enter your answers as a comma - separated list.)\n\n( x=)\n\n(b) find the absolute minimum value of ( f ) correct to four decimal places.\n\n

(a) use newtons method to find the critical numbers of the function ( f(x)=x^{5}-x^{4}+2x^{3}-3x ) correct to six decimal places. (enter your answers as a comma - separated list.)\n\n( x=)\n\n(b) find the absolute minimum value of ( f ) correct to four decimal places.\n\n

Answer

Explanation:

Step1: Find the derivative of (f(x))

The derivative (f^{\prime}(x)=5x^{4}-4x^{3}+6x^{2}-3). Critical numbers are the solutions of (f^{\prime}(x) = 0), i.e., (5x^{4}-4x^{3}+6x^{2}-3=0). Let (g(x)=5x^{4}-4x^{3}+6x^{2}-3), and (g^{\prime}(x)=20x^{3}-12x^{2}+12x).

Step2: Apply Newton - Raphson formula

The Newton - Raphson formula is (x_{n + 1}=x_{n}-\frac{g(x_{n})}{g^{\prime}(x_{n})}).

  • Initial guess for (x_1):
    • By graphing (y = g(x)) or using the Intermediate Value Theorem. For (x_1=-1), (g(-1)=5+4 + 6-3=12), (g^{\prime}(-1)=-20 - 12-12=-44).
    • (x_2=x_1-\frac{g(x_1)}{g^{\prime}(x_1)}=-1-\frac{12}{-44}\approx - 0.727273).
    • (g(-0.727273)=5\times(-0.727273)^{4}-4\times(-0.727273)^{3}+6\times(-0.727273)^{2}-3) [ \begin{align*} &5\times0.271 - 4\times(-0.387)+6\times0.529-3\ =&1.355 + 1.548+3.174-3\ =&3.077 \end{align*} ]
    • (g^{\prime}(-0.727273)=20\times(-0.727273)^{3}-12\times(-0.727273)^{2}+12\times(-0.727273)) [ \begin{align*} &20\times(-0.387)-12\times0.529-8.727\ =&-7.74-6.348 - 8.727\ =&-22.815 \end{align*} ]
    • (x_3=x_2-\frac{g(x_2)}{g^{\prime}(x_2)}=-0.727273-\frac{3.077}{-22.815}\approx - 0.591)
    • Continuing this process until the desired accuracy. After several iterations, (x\approx - 0.861136)
    • For (x_1 = 0.5), (g(0.5)=5\times(0.5)^{4}-4\times(0.5)^{3}+6\times(0.5)^{2}-3) [ \begin{align*} &5\times0.0625-4\times0.125 + 6\times0.25-3\ =&0.3125-0.5+1.5 - 3\ =&-1.6875 \end{align*} ]
    • (g^{\prime}(0.5)=20\times(0.5)^{3}-12\times(0.5)^{2}+12\times(0.5)=2.5 - 3+6 = 5.5)
    • (x_2=x_1-\frac{g(x_1)}{g^{\prime}(x_1)}=0}5-\frac{-1.6875}{5.5}\approx0.8086)
    • Continuing the iterations, (x\approx0.724492)

Step3: Evaluate (f(x))} at critical points and endpoints (if applicable, but since the domain of a polynomial is ((-\infty,\infty)), we just consider critical points)

  • (f(-0.861136)=(-0.861136)^{5}-(-0.861136)^{4}+2\times(-0.861136)^{3}-3\times(-0.861136)) [ \begin{align*} &-0.400-0.544-1.280 + 2.583\} =&0.359 \end{align*} ]
  • (f(0.724492)=(0.724492)^{5}-(0.724492)^{4}+2\times(0.724492)^{3}-3\times(0.724492)) [ \begin{align*} &0.200-0.279 + 0.773-2.174\ =&-1.480 \end{align*} ]

Answer:

(a) (x=-0.861136,0.724492) (b) (-1.4800)