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The sketch below shows the graph of g(x) = x³ − 6x − 1. P and Q are the turning points and R the y-intercept of g.

Mathematics

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6 parts
9.1
(0, -1)
9.2

(−2,42−1)(-\sqrt{2}, 4\sqrt{2} - 1)

(2,−42−1)(\sqrt{2}, -4\sqrt{2} - 1)

9.3

x∈(−∞,−2)∪(2,∞)x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)

9.4

x∈[−2,2]x \in [-\sqrt{2}, \sqrt{2}]

9.5

y=−6x−1y = -6x - 1

9.6

x=−2x = -\sqrt{2}

Step-by-step solution

14 steps

The given function is g(x)=x3−6x−1g(x) = x^3 - 6x - 1.

Part 9.1· 1 step

Determine the coordinates of R.

R is the y-intercept, which occurs when x=0x=0.

  1. Substitute x=0x=0 into the function g(x)g(x).

    g(0)=(0)3−6(0)−1g(0) = (0)^3 - 6(0) - 1 g(0)=−1g(0) = -1

    The coordinates of R are (0,−1)\boxed{(0, -1)}.

Part 9.2

(−2,42−1)(-\sqrt{2}, 4\sqrt{2} - 1)

Determine the coordinates of the turning points P and Q.

Turning points occur where the first derivative g′(x)g'(x) is equal to zero.

  1. Find the first derivative of g(x)g(x).

    g′(x)=ddx(x3−6x−1)=3x2−6g'(x) = \frac{d}{dx}(x^3 - 6x - 1) = 3x^2 - 6

  2. Set g′(x)=0g'(x) = 0 and solve for xx.

    3x2−6=03x^2 - 6 = 0 3x2=63x^2 = 6 x2=2x^2 = 2 x=±2x = \pm \sqrt{2}

  3. Substitute these xx values back into g(x)g(x) to find the corresponding yy values.

    For x=2x = \sqrt{2}:

    g(2)=(2)3−6(2)−1=22−62−1=−42−1g(\sqrt{2}) = (\sqrt{2})^3 - 6(\sqrt{2}) - 1 = 2\sqrt{2} - 6\sqrt{2} - 1 = -4\sqrt{2} - 1

    For x=−2x = -\sqrt{2}:

    g(−2)=(−2)3−6(−2)−1=−22+62−1=42−1g(-\sqrt{2}) = (-\sqrt{2})^3 - 6(-\sqrt{2}) - 1 = -2\sqrt{2} + 6\sqrt{2} - 1 = 4\sqrt{2} - 1

  4. Identify P and Q from the graph.

    P is the local maximum and Q is the local minimum.
    The coordinates of P are (−2,42−1)\boxed{(-\sqrt{2}, 4\sqrt{2} - 1)} and the coordinates of Q are (2,−42−1)\boxed{(\sqrt{2}, -4\sqrt{2} - 1)}.

Part 9.3

x∈(−∞,−2)∪(2,∞)x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)

Calculate the values of xx for which gg strictly increases as xx increases.

The function g(x)g(x) strictly increases when its first derivative g′(x)>0g'(x) > 0.

  1. Set g′(x)>0g'(x) > 0.

    3x2−6>03x^2 - 6 > 0 3x2>63x^2 > 6 x2>2x^2 > 2

  2. Solve the inequality for xx.

    x<−2orx>2x < -\sqrt{2} \quad \text{or} \quad x > \sqrt{2}

    The values of xx for which gg strictly increases are x∈(−∞,−2)∪(2,∞)\boxed{x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)}.

Part 9.4

x∈[−2,2]x \in [-\sqrt{2}, \sqrt{2}]

If h(x)=g′(x)h(x) = g'(x), determine for which values of xx is h(x)≤0h(x) \le 0.

This means finding the values of xx for which g′(x)≤0g'(x) \le 0.

  1. Set g′(x)≤0g'(x) \le 0.

    3x2−6≤03x^2 - 6 \le 0 3x2≤63x^2 \le 6 x2≤2x^2 \le 2

  2. Solve the inequality for xx.

    −2≤x≤2-\sqrt{2} \le x \le \sqrt{2}

    The values of xx for which h(x)≤0h(x) \le 0 are x∈[−2,2]\boxed{x \in [-\sqrt{2}, \sqrt{2}]}.

Part 9.5

y=−6x−1y = -6x - 1

Determine the equation of the tangent to gg at R.

The coordinates of R are (0,−1)(0, -1) from part 9.1. The gradient of the tangent at R is g′(0)g'(0).

  1. Calculate the gradient m=g′(0)m = g'(0).

    g′(x)=3x2−6g'(x) = 3x^2 - 6 m=g′(0)=3(0)2−6=−6m = g'(0) = 3(0)^2 - 6 = -6

  2. Use the point-slope form of a linear equation, y−y1=m(x−x1)y - y_1 = m(x - x_1).

    y−(−1)=−6(x−0)y - (-1) = -6(x - 0) y+1=−6xy + 1 = -6x y=−6x−1y = -6x - 1

    The equation of the tangent to gg at R is y=−6x−1\boxed{y = -6x - 1}.

Part 9.6

x=−2x = -\sqrt{2}

Write down the equation of the line perpendicular to the tangent at P.

The coordinates of P are (−2,42−1)(-\sqrt{2}, 4\sqrt{2} - 1) from part 9.2. P is a turning point, so the tangent at P is a horizontal line.

  1. The gradient of the tangent at P is g′(−2)=0g'(-\sqrt{2}) = 0.

  2. A line perpendicular to a horizontal line (gradient 0) is a vertical line.

  3. A vertical line passing through P (−2,42−1)(-\sqrt{2}, 4\sqrt{2} - 1) has the equation x=−2x = -\sqrt{2}.

    The equation of the line perpendicular to the tangent at P is x=−2\boxed{x = -\sqrt{2}}.

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