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Let \( f(x) = 2x^3 - 21x^2 + 40x + 63 \) where \( x \in \mathbb{R} \) - Leaving Cert Mathematics - Question b - 2022
Question b
Let \( f(x) = 2x^3 - 21x^2 + 40x + 63 \) where \( x \in \mathbb{R} \). (i) \( x + 1 \) is a factor of \( f(x) \). Find the three values of \( x \) for which \( f(x... show full transcript
Worked Solution & Example Answer:Let \( f(x) = 2x^3 - 21x^2 + 40x + 63 \) where \( x \in \mathbb{R} \) - Leaving Cert Mathematics - Question b - 2022
Step 1
Find the three values of \( x \) for which \( f(x) = 0 \)
Answer
To determine the values of ( x ) for which ( f(x) = 0 ), we will use the factorization method. Given that ( x + 1 ) is a factor of ( f(x) ), we will perform polynomial long division.
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Polynomial Division: Divide ( f(x) ) by ( x + 1 ):
( 2x^3 - 21x^2 + 40x + 63 ) ÷ ( (x + 1) ) yields:
( 2x^2 - 23x + 63 )Thus, we have:
( f(x) = (x + 1)(2x^2 - 23x + 63) ) -
Finding Roots of Quadratic: Now to find further zeros:
Set ( 2x^2 - 23x + 63 = 0 ) and apply the quadratic formula:
( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 2 \cdot 63}}{2 \cdot 2} )( = \frac{23 \pm \sqrt{529 - 504}}{4} = \frac{23 \pm \sqrt{25}}{4} )
( = \frac{23 \pm 5}{4} )Thus, the solutions are:
( x = \frac{28}{4} = 7 ), ( x = \frac{18}{4} = 4.5 ), and ( x = -1 ).Hence, the three values of ( x ) are: ( x = -1, \quad x = 4.5, \quad \text{or} \quad x = 7 ).
Step 2
Find the range of values of \( x \) for which \( f'(x) \) is negative, correct to 2 decimal places.
Answer
First, we need to find the derivative of ( f(x) ):
( f'(x) = 6x^2 - 42x + 40 ).
Next, we set ( f'(x) < 0 ) to find the range:
( 6x^2 - 42x + 40 < 0 ).
To solve this inequality, we can find the roots of the quadratic equation.
Using the quadratic formula:
( x = \frac{-(-42) \pm \sqrt{(-42)^2 - 4 \cdot 6 \cdot 40}}{2 \cdot 6} = \frac{42 \pm \sqrt{1764 - 960}}{12} = \frac{42 \pm \sqrt{804}}{12} )
Thus, ( x = \frac{42 \pm 28.4}{12} ) gives approximately:
( x = \frac{70.4}{12} \approx 5.87 )
and ( x = \frac{13.6}{12} \approx 1.13 ).
Therefore, the range of values of ( x ) for which ( f'(x) < 0 ) is:
( 1.14 < x < 5.86 ) (to 2 decimal places).