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Calculus, MET1 2024 VCAA 1b

Let  \(f(x)=\log _e\left(x^3-3 x+2\right)\).

Find  \(f^{\prime}(3)\)   (2 marks)

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\(\dfrac{6}{5}\)

Show Worked Solution

  \(f(x)\) \(=\log_{e}(x^3-3x+2)\)
  \(f^{\prime}(x)\) \(=\dfrac{3x^2-3}{x^3-3x+2}\)
  \(f^{\prime}(3)\) \(=\dfrac{3(3)^2-3}{(3)^3-3(3)+2}=\dfrac{6}{5}\)

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 VCAA 7 MC

Let  \(f(x)=\log_{e}x\), where  \(x>0\)  and  \(g(x)=\sqrt{1-x}\), where  \(x<1\).

The domain of the derivative of \((f\circ g)(x)\) is

  1. \(x\in R\)
  2. \(x\in (-\infty, 1]\)
  3. \(x\in (-\infty, 1)\)
  4. \(x\in (0, \infty)\)
  5. \(x\in (0, 1)\)
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\(C\)

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\(\text{Given }f(x)=\log_{e}x\ \ \text{and}\ \ g(x)=\sqrt{1-x}\)

\((f\circ g)(x)=\log_{e}\sqrt{1-x}=\dfrac{1}{2}\log_{e}(1-x)\)

\((f\circ g)^{′}(x)=\dfrac{1}{2}\times\dfrac{-1}{1-x}=\dfrac{1}{2(x-1)}\ \text{  where}\ \ x<1\)

\(\Rightarrow C\)

Calculus, MET1 2015 VCAA 1b

Let  `f(x) = (log_e(x))/(x^2)`.

  1. Find  `f^{prime}(x)`.   (2 marks)

  2. Evaluate  `f^{prime}(1)`.   (1 mark)

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  1. `(1 – 2log_e(x))/(x^3)`
  2. `1`
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i.   `text(Using Quotient Rule:)`

`(h/g)^{prime}` `= (h^{prime}g-hg^{prime})/(g^2)`
`f^{prime}(x)` `= ((1/x)x^2-log_e(x)*2x)/(x^4)`
  `= (1-2log_e(x))/(x^3)`

 

ii.    `f^{prime}(1)` `= (1-2log_e(1))/(1^3)`
    `= 1`

Calculus, MET2 2011 VCAA 4 MC

The derivative of  `log_e(2f(x))`  with respect to `x` is

  1. `(f′(x))/(f(x))`
  2. `2(f′(x))/(f(x))`
  3. `(f′(x))/(2f(x))`
  4. `log_e(2f′(x))`
  5. `2log_e(2f′(x))`
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`A`

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`text(Chain Rule:)`

`text(If)\ \ h(x)` `= f(g(x))`
`h′(x)` `= f′(g(x)) xx g′(x)`
`d/(dx)(log_e(2f(x)))` `= 1/(2f(x)) xx 2f′(x)`
  `= (f′(x))/(f(x))`

`=> A`

Calculus, MET1 2007 VCAA 2b

Let  `g(x) = log_e(tan(x))`.  Evaluate `g^{prime}(pi/4)`.   (2 marks)

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`g^{prime}(pi/4) = 2`

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`g(x) = log_e (tan(x))`

`text(Using Chain Rule:)`

`g^{prime}(x) = (sec^2(x))/(tan(x))`

`text(When)\ \ x = pi/4,`

`g^{prime}(pi/4)` `= (sec^2(pi/4))/(tan (pi/4))`
  `=1/(1/sqrt2)^2`
  `=1/(1/2)`
  `=2`