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IB Mathematics: Applications and Interpretation SL · 12.9: Differentiation · 12.9.12

Maximum and minimum points on a domain Problem set A Worked solutions

Skills

  1. For \(y=x^{3} - 12 x + 5\), \(-3\le x\le 5\), use your calculator to find (i) any local maximum and local minimum points, (ii) the greatest value of y.

    1. (a) Local maximum and minimum points.

      • Set the window to \(-3\le x\le 5\) and use the calculator's maximum and minimum tools
      • From the calculator: local maximum (−2, 21); local minimum (2, −11)

    2. (b) The greatest value of y.

      • End-points: x = −3 gives y = 14, x = 5 gives y = 70
      • Compare these with the local values. The largest is 70, at x = 5

  2. For \(y=2 x^{4} - 9 x^{2} + 1\), \(-2\le x\le 3\), use your calculator to find (i) any local maximum and local minimum points, (ii) the greatest value of y.

    1. (a) Local maximum and minimum points.

      • Set the window to \(-2\le x\le 3\) and use the calculator's maximum and minimum tools
      • From the calculator: local maximum (0, 1); local minimum (−1.50, −9.13) and (1.50, −9.13)

    2. (b) The greatest value of y.

      • End-points: x = −2 gives y = −3, x = 3 gives y = 82
      • Compare these with the local values. The largest is 82, at x = 3

  3. The curve \(y=x^{4} + b x^{2} + c\) has a horizontal tangent at the point (2, −5). Find the values of b and c.

    • \(\frac{dy}{dx}=4 x^{3} + 2 b x\)
    • At x = 2: \(4 b + 32=0\), so b = −8
    • The point is on the curve: \(c - 16=-5\)
    • c = 11

  4. Given that \(f(x)=x^{4} + x^{3} - 6 x + 2\), solve the equation

    1. (a) \(f(x)=0\);

      • Graph \(y=f(x)\)
      • From the calculator: the zeros of the graph
      • x = 0.342 or 1.38

    2. (b) \(f'(x)=0\).

      • Graph \(y=\frac{d}{dx}\left(x^{4} + x^{3} - 6 x + 2\right)\)
      • From the calculator: the zeros of the graph
      • x = 0.942

  5. Find the coordinates of the points on the graph of \(y=12x^{2}-x^{4}\) where the tangent is horizontal.

    • Graph \(y=\frac{d}{dx}\left(12x^{2}-x^{4}\right)\) and find its zeros: x = −2.45, 0, 2.45
    • Find y for each x by putting it into the function
    • (−2.45, 36), (0, 0), (2.45, 36)

  6. Find the coordinates of the local maximum point on the graph of \(y=6 x - \frac{8}{x^{2}}\).

    • Graph \(y=6 x - \frac{8}{x^{2}}\). Choose a window that shows the maximum, and use the calculator's maximum tool
    • From the calculator: x = −1.39, y = −12.5
    • Give both coordinates
    • (−1.39, −12.5)

  7. A function is defined by \(f(x)=3 x^{3} - 7 x^{2} - 5 x + 6\).

    1. (a) Find the coordinates of the local minimum point on the graph of \(y=f(x)\).

      • Graph \(y=f(x)\) and use the calculator's minimum tool
      • From the calculator: local minimum (1.86, −8.21)

    2. (b) Find the smallest value of the function on the domain \(-2\le x\le 3\).

      • Check the end-points: x = −2 gives −36, x = 3 gives 9
      • Compare with the local minimum value −8.21. The smallest is −36

  8. Find the range of each function.

    1. (a) \(f(x)=4 x + \frac{9}{x}\), for \(1\le x\le 4\);

      • Graph \(y=f(x)\) for \(1\le x\le 4\)
      • From the calculator: local minimum (1.50, 12)
      • End-points: x = 1 gives 13, x = 4 gives 18.3
      • Smallest value 12, greatest value 18.3

    2. (b) \(f(x)=3 x^{4} - 8 x^{3} - 6 x^{2} + 24 x\), for all real x.

      • Graph \(y=f(x)\). The graph rises without limit as x gets large
      • From the calculator: local minima at (−1, −19), (2, 8)
      • The lowest point of the graph has y = −19

  9. Let \(f(x)=x^{3} - 9 x^{2} + 15 x + 4\).

    1. (a) Use your calculator to solve \(f'(x)=0\).

      • From the calculator: graph \(f'(x)\) and read its zeros
      • x = 1 and x = 5

    2. (b) Write down the interval on which f is decreasing.

      • The graph of \(f'\) is above the axis for x < 1, below it between 1 and 5, above it for x > 5
      • f is decreasing for \(1<x<5\)

    3. (c) Find the local maximum point and the local minimum point of the graph of f.

      • At x = 1, \(f'\) goes from positive to negative: local maximum
      • At x = 5, \(f'\) goes from negative to positive: local minimum
      • \(f(1)=11\) and \(f(5)=-21\)

  10. The curve \(y=x^{3} - 3 x^{2} + 3 x + 4\) has a horizontal tangent at the point \((1, 5)\).

    1. (a) Show that the gradient of the curve is 0 at \(x=1\).

      • \(\frac{dy}{dx}=3 x^{2} - 6 x + 3=3 (x - 1)^{2}\)
      • At \(x=1\) this is 0

    2. (b) Find the gradient at \(x=0\) and at \(x=2\).

      • x = 0: \(\frac{dy}{dx}=3\)
      • x = 2: \(\frac{dy}{dx}=3\)

    3. (c) State whether \((1, 5)\) is a local maximum, a local minimum or neither. Give a reason.

      • The gradient is positive on both sides of the point
      • It does not change sign, so the point is neither a local maximum nor a local minimum

In context

  1. The temperature, \(T\) °C, in a greenhouse is modelled by \(T=0.02t^3-0.7t^2+6t+8\), where t is the time in hours after midnight, \(0\le t\le 24\). Use your calculator to find

    1. (a) the highest temperature, and when it happens;

      • Graph the model for \(0\le t\le 24\)
      • From the calculator: local maximum (5.66, 23.2), local minimum (17.7, 5.80)
      • End-points: t = 0 gives 8; t = 24 gives 25.3
      • The greatest value is 25.3, at the end-point t = 24

    2. (b) the lowest temperature, and when it happens.

      • Use the same graph and the same end-points
      • The smallest of the end-point values and the local minimum 5.80 is 5.80
      • It occurs at the local point t = 17.7

  2. The height of a roller-coaster track, \(h\) metres, at time t seconds is modelled by \(h=at^3+bt^2-12t+30\). The track has a local minimum at the point (2, 10). Find the values of a and b.

    • \(\frac{dh}{dt}=3 a t^{2} + 2 b t - 12\)
    • Horizontal tangent at t = 2: \(12 a - 12 + 4 b=0\)
    • The point is on the curve: \(8 a - 4 + 4 b=0\)
    • Solve the two equations together on the calculator
    • a = 2, b = −3

  3. The depth of water in a reservoir, h metres, t months after 1 January is modelled by \(h=0.1t^3-1.8t^2+8.1t+20\), for \(0\le t\le 12\).

    1. (a) Find the values of t for which \(\frac{dh}{dt}=0\).

      • \(\frac{dh}{dt}=0.3t^2-3.6t+8.1\)
      • From the calculator: zeros at t = 3 and t = 9

    2. (b) Find the months during which the depth is falling.

      • The graph of \(\frac{dh}{dt}\) is below the axis between t = 3 and t = 9
      • The depth is falling for \(3<t<9\)

    3. (c) The depth has a local maximum and a local minimum in this model. Find the time and the depth of each.

      • At t = 3 the gradient goes from positive to negative: local maximum, \(h(3)=30.8\)
      • At t = 9 the gradient goes from negative to positive: local minimum, \(h(9)=20\)

Stretch

  1. The graph of \(y=x^{4} - 4 x^{3} - 8 x^{2} + 48 x\) has three points where the tangent is horizontal.

    1. (a) Find the coordinates of the three points.

      • Graph \(y=\frac{d}{dx}\left(x^{4} - 4 x^{3} - 8 x^{2} + 48 x\right)\)
      • From the calculator: zeros at x = −2, 2, 3
      • Put each x into f(x) to find y: (−2, −80), (2, 48), (3, 45)

    2. (b) Write down the range of the function \(f(x)=x^{4} - 4 x^{3} - 8 x^{2} + 48 x\) for all real x.

      • The graph rises without limit as x gets large in either direction
      • Compare the y-values of the local minimum points
      • The lowest point has y = −80, so \(f(x)\ge -80\)

  2. The curve \(y=ax^2+\frac{b}{x}\) has a horizontal tangent at the point (2, 6). Find the values of a and b.

    • \(\frac{dy}{dx}=2 a x - \frac{b}{x^{2}}\)
    • Horizontal tangent at x = 2: \(4 a - \frac{b}{4}=0\)
    • The point is on the curve: \(4 a - 6 + \frac{b}{2}=0\)
    • a = 0.5, b = 8

jmaths.xyz · AI SL · 12.9.12 Maximum and minimum points on a domain