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Calculus, 2ADV C3 2024 MET2 3*

A function is defined as  \(f(x)=\dfrac{x-h}{(x+1)(x-4)}\)

Determine the range of \(h\) where the graph of \(f(x)\) have no turning points.   (3 marks)

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\(-1 \leq h \leq 4\)

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\(f(x)=\dfrac{x-h}{(x+1)(x-4)}\)

\(\text{Let}\ \ u=x-h\ \ \Rightarrow\ \ u^{′} = 1\)

 \(v=x^2-3x-4\ \ \Rightarrow\ \ v^{′} = 2x-3\)

\(f^{′}(x)=\dfrac{(x^2-3x-4)-(2x-3)(x-h)}{(x+1)^2(x-4)^2}\)

\(\text{Solve}\ \ f^{′}(x)=0:\)

\(x^2-3x-4-2x^2+2xh+3x-3h\) \(=0\)  
\(-x^2+2xh-(4+3h)\) \(=0\)  
\(x^2-2xh+(4+3h)\) \(=0\)  
\(x\) \(= \dfrac{2h \pm \sqrt{4h^2-4(4+3h)}}{2}\)  
  \(=h \pm \sqrt{h^2-3h-4}\)  

 
\(\text{No turning points occur if}\ \ h^2-3h-4=(h-4)(h+1)<0\)

\(-1 \lt h \lt 4\)

\(\text{If}\ \ h=-1\ \ \text{or}\ \ 4, f(x)\ \text{is linear (no TPs)}\)

\(\therefore -1 \leq h \leq 4\)

Calculus, 2ADV C3 2018 HSC 14c

Let  `f(x) = x^3 + kx^2 + 3x - 5`, where `k` is a constant.

Find the values of `k` for which  `f(x)`  has NO stationary points.  (3 marks)

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`-3 < k < 3`

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`f(x) = x^3 + kx^2 + 3x – 5`

♦ Mean mark 49%.

`f prime(x) = 3x^2 + 2kx + 3`
 

`text(No S.P.’s exist if)\ \ f prime(x)\ \ text(has no roots,)`

`Delta` `< 0`
`b^2 – 4ac` `< 0`
`(2k)^2 – 4 xx 3 xx 3` `< 0`
`4k^2 – 36` `< 0`
`k^2 – 9` `< 0`
`(k – 3) (k + 3)` `< 0`

 

`:. -3 < k < 3`