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If f(x) = -πx³, determine f'(x) from the first principles. Determine dy/dx if y = (1 − √x)³. Determine g'(x) if g(x) = x(x² - 100) - x² + 100 / x - 1. If h(x) = 7 − 2/x², calculate the value of h'(2) + 4.

Mathematics

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6 parts
7.1

f′(x)=−2πxf'(x) = -2\pi x

7.2

dydx=1−1x\frac{dy}{dx} = 1 - \frac{1}{\sqrt{x}}

7.3

g′(x)=2xg'(x) = 2x

7.4

h′(2)+4=92h'(2)+4 = \frac{9}{2}

8.1
Shown:

d=3d=3 and q=2q=2

8.2

C(1;4)

Step-by-step solution

20 steps

Here are the solutions to Questions 7 and 8.

Part 7

Part 7.1

f′(x)=−2πxf'(x) = -2\pi x

Determine f′(x)f'(x) from the first principles if f(x)=−πx2f(x) = -\pi x^2.

  1. Write down the formula for differentiation from first principles.

    f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
  2. Substitute f(x)=−πx2f(x) = -\pi x^2 and f(x+h)=−π(x+h)2f(x+h) = -\pi (x+h)^2 into the formula.

    f′(x)=lim⁡h→0−π(x+h)2−(−πx2)hf'(x) = \lim_{h \to 0} \frac{-\pi (x+h)^2 - (-\pi x^2)}{h} f′(x)=lim⁡h→0−π(x2+2xh+h2)+πx2hf'(x) = \lim_{h \to 0} \frac{-\pi (x^2 + 2xh + h^2) + \pi x^2}{h} f′(x)=lim⁡h→0−πx2−2πxh−πh2+πx2hf'(x) = \lim_{h \to 0} \frac{-\pi x^2 - 2\pi xh - \pi h^2 + \pi x^2}{h} f′(x)=lim⁡h→0−2πxh−πh2hf'(x) = \lim_{h \to 0} \frac{-2\pi xh - \pi h^2}{h}
  3. Factor out hh from the numerator and simplify.

    f′(x)=lim⁡h→0h(−2πx−πh)hf'(x) = \lim_{h \to 0} \frac{h(-2\pi x - \pi h)}{h} f′(x)=lim⁡h→0(−2πx−πh)f'(x) = \lim_{h \to 0} (-2\pi x - \pi h)
  4. Evaluate the limit as h→0h \to 0.

    f′(x)=−2πx−π(0)f'(x) = -2\pi x - \pi (0) f′(x)=−2πxf'(x) = -2\pi x f′(x)=−2πx\boxed{f'(x) = -2\pi x}

Part 7.2

dydx=1−1x\frac{dy}{dx} = 1 - \frac{1}{\sqrt{x}}

Determine dydx\frac{dy}{dx} if y=(1−x)2y = (1-\sqrt{x})^2.

  1. Expand the expression for yy.

    y=(1−x1/2)2y = (1 - x^{1/2})^2 y=1−2x1/2+(x1/2)2y = 1 - 2x^{1/2} + (x^{1/2})^2 y=1−2x1/2+xy = 1 - 2x^{1/2} + x
  2. Differentiate each term with respect to xx.

    dydx=ddx(1)−ddx(2x1/2)+ddx(x)\frac{dy}{dx} = \frac{d}{dx}(1) - \frac{d}{dx}(2x^{1/2}) + \frac{d}{dx}(x) dydx=0−2(12x12−1)+1\frac{dy}{dx} = 0 - 2 \left(\frac{1}{2}x^{\frac{1}{2}-1}\right) + 1 dydx=−x−1/2+1\frac{dy}{dx} = -x^{-1/2} + 1 dydx=−1x+1\frac{dy}{dx} = -\frac{1}{\sqrt{x}} + 1 dydx=1−1x\boxed{\frac{dy}{dx} = 1 - \frac{1}{\sqrt{x}}}

Part 7.3

g′(x)=2xg'(x) = 2x

Determine g′(x)g'(x) if g(x)=x(x2−100)−x2+100x−1g(x) = \frac{x(x^2-100)-x^2+100}{x-1}.

  1. Simplify the expression for g(x)g(x) by factoring the numerator.

    g(x)=x(x2−100)−(x2−100)x−1g(x) = \frac{x(x^2-100) - (x^2-100)}{x-1} g(x)=(x−1)(x2−100)x−1g(x) = \frac{(x-1)(x^2-100)}{x-1}

    For x≠1x \neq 1, we can cancel the (x−1)(x-1) terms:

    g(x)=x2−100g(x) = x^2 - 100
  2. Differentiate g(x)=x2−100g(x) = x^2 - 100 with respect to xx.

    g′(x)=ddx(x2)−ddx(100)g'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(100) g′(x)=2x−0g'(x) = 2x - 0 g′(x)=2xg'(x) = 2x g′(x)=2x\boxed{g'(x) = 2x}

Part 7.4

h′(2)+4=92h'(2)+4 = \frac{9}{2}

If h(x)=7−2x2h(x) = 7 - \frac{2}{x^2}, calculate the value of h′(2)+4h'(2)+4.

  1. Rewrite h(x)h(x) using negative exponents.

    h(x)=7−2x−2h(x) = 7 - 2x^{-2}
  2. Determine h′(x)h'(x).

    h′(x)=ddx(7)−ddx(2x−2)h'(x) = \frac{d}{dx}(7) - \frac{d}{dx}(2x^{-2}) h′(x)=0−2(−2x−2−1)h'(x) = 0 - 2(-2x^{-2-1}) h′(x)=4x−3h'(x) = 4x^{-3} h′(x)=4x3h'(x) = \frac{4}{x^3}
  3. Calculate h′(2)h'(2).

    h′(2)=4(2)3h'(2) = \frac{4}{(2)^3} h′(2)=48h'(2) = \frac{4}{8} h′(2)=12h'(2) = \frac{1}{2}
  4. Calculate h′(2)+4h'(2)+4.

    h′(2)+4=12+4h'(2)+4 = \frac{1}{2} + 4 h′(2)+4=12+82h'(2)+4 = \frac{1}{2} + \frac{8}{2} h′(2)+4=92h'(2)+4 = \frac{9}{2} h′(2)+4=92\boxed{h'(2)+4 = \frac{9}{2}}

Part 8

The graph below represents the function f(x)=−x3+dx+qf(x) = -x^3 + dx + q with A(−1;0)A(-1;0) and B(2;0)B(2;0).

Part 8.1

d=3d=3 and q=2q=2

Show that d=3d=3 and q=2q=2.

  1. Use point A(−1;0)A(-1;0) to form an equation.

    Substitute x=−1x=-1 and f(x)=0f(x)=0 into f(x)=−x3+dx+qf(x) = -x^3 + dx + q:

    0=−(−1)3+d(−1)+q0 = -(-1)^3 + d(-1) + q 0=−(−1)−d+q0 = -(-1) - d + q 0=1−d+q0 = 1 - d + q d−q=1(Equation 1)d - q = 1 \quad \text{(Equation 1)}
  2. Use point B(2;0)B(2;0) to form another equation.

    Substitute x=2x=2 and f(x)=0f(x)=0 into f(x)=−x3+dx+qf(x) = -x^3 + dx + q:

    0=−(2)3+d(2)+q0 = -(2)^3 + d(2) + q 0=−8+2d+q0 = -8 + 2d + q 2d+q=8(Equation 2)2d + q = 8 \quad \text{(Equation 2)}
  3. Solve the system of equations.

    Add Equation 1 and Equation 2:

    (d−q)+(2d+q)=1+8(d - q) + (2d + q) = 1 + 8 3d=93d = 9 d=3d = 3
  4. Substitute d=3d=3 into Equation 1 to find qq.

    3−q=13 - q = 1 −q=1−3-q = 1 - 3 −q=−2-q = -2 q=2q = 2

    Therefore, d=3d=3 and q=2q=2.

    d=3 and q=2\boxed{d=3 \text{ and } q=2}

Part 8.2C(1;4)

Determine the coordinates of turning point C.

  1. Write the full equation for f(x)f(x) using d=3d=3 and q=2q=2.

    f(x)=−x3+3x+2f(x) = -x^3 + 3x + 2
  2. Find the derivative f′(x)f'(x) to locate turning points.

    f′(x)=ddx(−x3+3x+2)f'(x) = \frac{d}{dx}(-x^3 + 3x + 2) f′(x)=−3x2+3f'(x) = -3x^2 + 3
  3. Set f′(x)=0f'(x)=0 to find the x-coordinates of the turning points.

    −3x2+3=0-3x^2 + 3 = 0 −3x2=−3-3x^2 = -3 x2=1x^2 = 1 x=±1x = \pm 1

    The turning points occur at x=−1x=-1 and x=1x=1. From the graph, point C is the turning point with a positive x-coordinate, so xC=1x_C = 1.

  4. Substitute xC=1x_C=1 into f(x)f(x) to find the y-coordinate of C.

    f(1)=−(1)3+3(1)+2f(1) = -(1)^3 + 3(1) + 2 f(1)=−1+3+2f(1) = -1 + 3 + 2 f(1)=4f(1) = 4

    The coordinates of turning point C are (1;4)(1;4).

    C(1;4)\boxed{C(1;4)}
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