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Part of the graph of a cubic function $f(x)$ is shown below (graph not to scale) - Leaving Cert Mathematics - Question 4 (b) - 2019

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Question 4 (b)

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Part of the graph of a cubic function $f(x)$ is shown below (graph not to scale). The graph cuts the x-axis at the three points A(2, 0), B, and C. (i) Given that $f... show full transcript

Worked Solution & Example Answer:Part of the graph of a cubic function $f(x)$ is shown below (graph not to scale) - Leaving Cert Mathematics - Question 4 (b) - 2019

Step 1

Given that $f'(x) = 6x^2 - 54x + 109$, show that $f(x) = 2x^3 - 27x^2 + 109x - 126$.

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Answer

To find f(x)f(x), we need to integrate the derivative:

  1. Integrate f′(x)f'(x): f(x) &= \\int (6x^2 - 54x + 109) \, dx \\ \\\ \\ \\ \\ \\ \\ \\\

\ (x) = 2x^3 - 27x^2 + 109x + C dl\
\end{align*}$$

  1. Evaluate CC using the point A(2, 0):
    • Substitute x=2x = 2 into f(x)f(x):
    f(2) &= 2(2)^3 - 27(2)^2 + 109(2) + C = 0 \\ &= 16 - 108 + 218 + C = 0 \\[\] = 126 + C = 0 \\[ \\]

C = -126 \[] \end{align*}$$

  1. Thus, we conclude: f(x)=2x3−27x2+109x−126f(x) = 2x^3 - 27x^2 + 109x - 126

Step 2

Find the co-ordinates of the point B and the point C.

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Answer

To find points B and C, we need to solve for the roots of f(x)=0f(x) = 0:

  1. We have the equation: 2x3−27x2+109x−126=02x^3 - 27x^2 + 109x - 126 = 0

    • Since we know x=2x = 2 is a root, we can factor it: (x−2)(2x2−23x+63)=0(x - 2)(2x^2 - 23x + 63) = 0

  2. Now find the roots of the quadratic 2x2−23x+63=02x^2 - 23x + 63 = 0 using the quadratic formula: x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    • Here, a=2a = 2, b=−23b = -23, c=63c = 63: x=23±(−23)2−4⋅2⋅632⋅2x = \frac{23 \pm \sqrt{(-23)^2 - 4 \cdot 2 \cdot 63}}{2 \cdot 2} =23±529−5044= \frac{23 \pm \sqrt{529 - 504}}{4} =23±254= \frac{23 \pm \sqrt{25}}{4} =23±54= \frac{23 \pm 5}{4}

    Thus, the solutions are:

    • x=284=7x = \frac{28}{4} = 7 (Point C)
    • x=184=4.5x = \frac{18}{4} = 4.5 (Point B)

  3. Finally, substitute xx values back into f(x)f(x) to find corresponding yy values:

    • For point B (4.5): f(4.5)=2(4.5)3−27(4.5)2+109(4.5)−126=0f(4.5) = 2(4.5)^3 - 27(4.5)^2 + 109(4.5) - 126 = 0
    • For point C (7): f(7)=2(7)3−27(7)2+109(7)−126=0f(7) = 2(7)^3 - 27(7)^2 + 109(7) - 126 = 0

  4. Thus the coordinates are:

    • B(4.5,0)B(4.5, 0)
    • C(7,0)C(7, 0)

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