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f(x) = x^4 - 4x - 8 - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6

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f(x)-=-x^4---4x---8-Edexcel-A-Level Maths Pure-Question 7-2007-Paper 6.png

f(x) = x^4 - 4x - 8. (a) Show that there is a root of f(x) = 0 in the interval [-2, -1]. (b) Find the coordinates of the turning point on the graph of y = f(x). (... show full transcript

Worked Solution & Example Answer:f(x) = x^4 - 4x - 8 - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 6

Step 1

Show that there is a root of f(x) = 0 in the interval [-2, -1]

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Answer

To show that there is a root in the interval [-2, -1], we evaluate f(-2) and f(-1):

  1. Calculate f(-2):

    f(2)=(2)44(2)8=16+88=16>0f(-2) = (-2)^4 - 4(-2) - 8 = 16 + 8 - 8 = 16 > 0
  2. Calculate f(-1):

    f(1)=(1)44(1)8=1+48=3<0f(-1) = (-1)^4 - 4(-1) - 8 = 1 + 4 - 8 = -3 < 0

Since f(-2) > 0 and f(-1) < 0, by the Intermediate Value Theorem, there is at least one root in the interval [-2, -1].

Step 2

Find the coordinates of the turning point on the graph of y = f(x)

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Answer

First, we find the derivative of f(x):

ddxf(x)=4x34\frac{d}{dx} f(x) = 4x^3 - 4

Setting the derivative equal to zero to find turning points:

4x34=0x3=1x=14x^3 - 4 = 0 \Rightarrow x^3 = 1 \Rightarrow x = 1

Now we substitute x = 1 back into f(x) to find the coordinates:

f(1)=144(1)8=148=11f(1) = 1^4 - 4(1) - 8 = 1 - 4 - 8 = -11

Thus, the coordinates of the turning point are (1, -11).

Step 3

Given that f(x) = (x - 2)(x^2 + ax^2 + bx + c), find the values of the constants, a, b, and c

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Answer

Expanding the expression:

f(x)=(x2)(x2+ax2+bx+c)=x3+(a2)x2+(b2a)x2cf(x) = (x - 2)(x^2 + ax^2 + bx + c) = x^3 + (a-2)x^2 + (b-2a)x - 2c

From f(x) = x^4 - 4x - 8, we can equate coefficients:

  1. Coefficient of x3x^3: a - 2 = 0
    a=2a = 2

  2. Coefficient of x2x^2: b - 2a = 0
    b=4b = 4

  3. Constant term: -2c = -8
    c=4c = 4

Thus, the values are: a = 2, b = 4, c = 4.

Step 4

In the space provided on page 21, sketch the graph of y = f(x)

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Answer

To sketch the graph of y = f(x), we plot the turning point at (1, -11) and identify the behavior as x approaches ±∞. The graph will approach +∞ as x approaches ±∞ due to the positive leading coefficient. It intersects the y-axis at f(0) = -8. The overall shape resembles a quartic polynomial with one local maximum and one local minimum.

Step 5

Hence sketch the graph of y = |f(x)|

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The graph of y = |f(x)| will reflect any part of the graph of y = f(x) that is below the x-axis. Starting from the turning point, the portions of the graph below the x-axis will be flipped above the x-axis. Conventional points where f(x) ≤ 0 will be shown on the upper half of the graph, creating a 'U' shape for those areas, while all other parts remain unchanged.

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