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Calculus, MET1 2023 VCAA SM-Bank 6

Newton's method is used to estimate the \(x\)-intercept of the function  \(f(x)=\dfrac{1}{3} x^3+2 x+4\).

  1. Verify that \(f(-1)>0\) and \(f(-2)<0\).  (1 mark)

  2. Using an initial estimate of \(x_0=-1\), find the value of \(x_1\).  (2 marks)

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a.    \(\text{See worked solutions}\)

b.    \(-\dfrac{14}{9}\)

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{1}{3}x^3+2x+4\)
  \(f(-1)\) \(=\dfrac{1}{3}.(-1)^3+2.(-1)+4=\dfrac{5}{3}>0\)
  \(f(-2)\) \(=\dfrac{1}{3}.(-2)^3+2.(-2)+4=-\dfrac{8}{3}<0\)

 

b.     \(x_1\) \(=x_0-\dfrac{f(x_0)}{f^{\prime}(x_0)},\quad\ \text{where }x_0=-1\)
   

\(=-1-\dfrac{-\dfrac{1}{3}+2.(-1)+4}{(-1)^2+2}\)
    \(=-\dfrac{14}{9}\)

Calculus, MET2 2023 SM-Bank 1

The function \(g\) is defined as follows.

\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)

  1. Sketch the graph of \(g\) on the axes below. Label the vertical asymptote with its equation, and label any axial intercepts, stationary points and endpoints in coordinate form, correct to three decimal places.   (3 marks)

     

     

  2.  i. Find the equation of the tangent to the graph of \(g\) at the point where \(x=1\).   (1 mark)

  3. ii. Sketch the graph of the tangent to the graph of \(g\) at \(x=1\) on the axes in part a.   (1 mark)

Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).

  1. Find the value of \(x_1\).   (1 mark)

  2. Find the horizontal distance between \(x_3\) and the closest \(x\)-intercept of \(g\), correct to four decimal places.   (1 mark)

  3.  i. Find the value of \(k\), where \(k>1\), such that an initial estimate of  \(x_0=k\)  gives the same value of  \(x_1\)  as found in part \(c\). Give your answer correct to three decimal places.   (2 marks)

  4. ii. Using this value of \(k\), sketch the tangent to the graph of \(g\) at the point where  \(x=k\)  on the axes in part a.   (1 mark)

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a.    
  
b.i

 \(y=2x-3\)
  
b.ii
  
c. \(\dfrac{3}{2}=1.5\)
  
d. \(0.0036\)
  
e.i \(k=2.397\)
  
e.ii
Show Worked Solution

a.

b.i    \(g(x)\) \(=3\log_{e}x-x\)
  \(g(1)\) \(=3\log_{e}1-1=-1\)
  \(g^{\prime}(x)\) \(=\dfrac{3}{x}-1\)
  \(g^{\prime}(1)\) \(=\dfrac{3}{1}-1=2\)

  
\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)

\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)

b.ii

 
c.    \(\text{Newton’s Method}\)

\(x_1\) \(=x_0-\dfrac{g(x)}{g'(x)}\)
  \(=1-\left(\dfrac{-1}{2}\right)\)
  \(=\dfrac{3}{2}=1.5\)

\(\text{Using CAS:}\)
  
 

d.    \(\text{Using CAS}\)

\(x\text{-intercept}:\ x=1.85718\)

\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)

 

e.i.  \(\text{Using CAS}\)

\(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\) \(=1.5\)
\(k>1\ \therefore\ \ k\) \(=2.397\)

 
e.ii

Calculus, MET2 2023 VCE SM-Bank 7 MC

One way of implementing Newton's method using pseudocode, with a tolerance level of 0.001 , is shown below.

The pseudocode is incomplete, with two missing lines indicated by an empty box.
 

  

Which one of the following options would be most appropriate to fill the empty box?
 



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\(E\)

Show Worked Solution

\(\text{The tolerance}=\pm 0.001\)

\(\Big|\text{next_ x}-\text{prev_ x}\Big|<\ \text{tolerance}\)

\(\therefore\ \textbf{If }\ \ -0.001<\ \text{next_ x}-\text{prev_ x}\ <0.001\ \textbf{Then }\)

\(\text{If true }\textbf{Return }\text{next_ x}\)

\(\Rightarrow E\)