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

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

Skills

  1. For \(y=x^{3} - 27 x + 4\), \(-4\le x\le 7\), 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 \(-4\le x\le 7\) and use the calculator's maximum and minimum tools
      • From the calculator: local maximum (−3, 58); local minimum (3, −50)

    2. (b) The greatest value of y.

      • End-points: x = −4 gives y = 48, x = 7 gives y = 158
      • Compare these with the local values. The largest is 158, at x = 7

  2. For \(y=3 x^{4} - 12 x^{2} + 5\), \(-1\le x\le 2.5\), 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 \(-1\le x\le 2.5\) and use the calculator's maximum and minimum tools
      • From the calculator: local maximum (0, 5); local minimum (1.41, −7)

    2. (b) The greatest value of y.

      • End-points: x = −1 gives y = −4, x = 2.50 gives y = 47.2
      • Compare these with the local values. The largest is 47.2, at x = 2.50

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

    • \(\frac{dy}{dx}=4 x^{3} + 2 b x\)
    • At x = 3: \(6 b + 108=0\), so b = −18
    • The point is on the curve: \(c - 81=4\)
    • c = 85

  4. Given that \(f(x)=x^{4} - x^{3} - 5 x + 1\), solve the equation

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

      • Graph \(y=f(x)\)
      • From the calculator: the zeros of the graph
      • x = 0.199 or 2.06

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

      • Graph \(y=\frac{d}{dx}\left(x^{4} - x^{3} - 5 x + 1\right)\)
      • From the calculator: the zeros of the graph
      • x = 1.39

  5. Find the coordinates of the points on the graph of \(y=8x^{2}-x^{4}\) where the tangent is horizontal.

    • Graph \(y=\frac{d}{dx}\left(8x^{2}-x^{4}\right)\) and find its zeros: x = −2, 0, 2
    • Find y for each x by putting it into the function
    • (−2, 16), (0, 0), (2, 16)

  6. Find the coordinates of the local maximum point on the graph of \(y=4 x - \frac{5}{x^{2}}\).

    • Graph \(y=4 x - \frac{5}{x^{2}}\). Choose a window that shows the maximum, and use the calculator's maximum tool
    • From the calculator: x = −1.36, y = −8.14
    • Give both coordinates
    • (−1.36, −8.14)

  7. A function is defined by \(f(x)=2 x^{3} - 5 x^{2} - 4 x + 3\).

    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 (2, −9)

    2. (b) Find the smallest value of the function on the domain \(-2\le x\le 4\).

      • Check the end-points: x = −2 gives −25, x = 4 gives 35
      • Compare with the local minimum value −9. The smallest is −25

  8. Find the range of each function.

    1. (a) \(f(x)=5 x + \frac{20}{x}\), for \(1\le x\le 6\);

      • Graph \(y=f(x)\) for \(1\le x\le 6\)
      • From the calculator: local minimum (2, 20)
      • End-points: x = 1 gives 25, x = 6 gives 33.3
      • Smallest value 20, greatest value 33.3

    2. (b) \(f(x)=3 x^{4} - 16 x^{3} + 18 x^{2}\), for all real x.

      • Graph \(y=f(x)\). The graph rises without limit as x gets large
      • From the calculator: local minima at (0, 0), (3, −27)
      • The lowest point of the graph has y = −27

  9. Let \(g(x)=x^{3} - 12 x^{2} + 36 x + 7\).

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

      • From the calculator: graph \(g'(x)\) and read its zeros
      • x = 2 and x = 6

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

      • The graph of \(g'\) is above the axis for x < 2, below it between 2 and 6, above it for x > 6
      • g is decreasing for \(2<x<6\)

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

      • At x = 2, \(g'\) goes from positive to negative: local maximum
      • At x = 6, \(g'\) goes from negative to positive: local minimum
      • \(g(2)=39\) and \(g(6)=7\)

  10. The curve \(y=x^{3} + 6 x^{2} + 12 x - 2\) has a horizontal tangent at the point \((-2, -10)\).

    1. (a) Show that the gradient of the curve is 0 at \(x=-2\).

      • \(\frac{dy}{dx}=3 x^{2} + 12 x + 12=3 (x + 2)^{2}\)
      • At \(x=-2\) this is 0

    2. (b) Find the gradient at \(x=-3\) and at \(x=-1\).

      • x = -3: \(\frac{dy}{dx}=3\)
      • x = -1: \(\frac{dy}{dx}=3\)

    3. (c) State whether \((-2, -10)\) 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 electricity demand, \(D\) megawatts, of a town is modelled by \(D=0.5t^3-14t^2+110t+20\), where t is the time in hours after 6 am, \(0\le t\le 18\). Use your calculator to find

    1. (a) the greatest demand, and when it happens;

      • Graph the model for \(0\le t\le 18\)
      • From the calculator: local maximum (5.62, 285), local minimum (13.0, 182)
      • End-points: t = 0 gives 20; t = 18 gives 380
      • The greatest value is 380, at the end-point t = 18

    2. (b) the least demand, and when it happens.

      • Use the same graph and the same end-points
      • The smallest of the end-point values and the local minimum 182 is 20
      • It occurs at the end-point t = 0

  2. The depth of a river bed, \(h\) metres below a bridge, at distance t hundred metres from the bank is modelled by \(h=at^3+bt^2-9t+31\). The depth has a local minimum at the point (3, 4). Find the values of a and b.

    • \(\frac{dh}{dt}=3 a t^{2} + 2 b t - 9\)
    • Horizontal tangent at t = 3: \(27 a - 9 + 6 b=0\)
    • The point is on the curve: \(27 a + 9 b=0\)
    • Solve the two equations together on the calculator
    • a = 1, b = −3

  3. The monthly profit of a shop, P thousand euros, t months after it opens is modelled by \(P=0.1t^3-1.2t^2+3.6t+5\), for \(0\le t\le 8\).

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

      • \(\frac{dP}{dt}=0.3t^2-2.4t+3.6\)
      • From the calculator: zeros at t = 2 and t = 6

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

      • The graph of \(\frac{dP}{dt}\) is below the axis between t = 2 and t = 6
      • The profit is falling for \(2<t<6\)

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

      • At t = 2 the gradient goes from positive to negative: local maximum, \(P(2)=8.2\)
      • At t = 6 the gradient goes from negative to positive: local minimum, \(P(6)=5\)

Stretch

  1. The graph of \(y=3 x^{4} - 20 x^{3} + 12 x^{2} + 96 x\) has three points where the tangent is horizontal.

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

      • Graph \(y=\frac{d}{dx}\left(3 x^{4} - 20 x^{3} + 12 x^{2} + 96 x\right)\)
      • From the calculator: zeros at x = −1, 2, 4
      • Put each x into f(x) to find y: (−1, −61), (2, 128), (4, 64)

    2. (b) Write down the range of the function \(f(x)=3 x^{4} - 20 x^{3} + 12 x^{2} + 96 x\) 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 = −61, so \(f(x)\ge -61\)

  2. The curve \(y=ax^2+\frac{b}{x}\) has a horizontal tangent at the point (1, 12). Find the values of a and b.

    • \(\frac{dy}{dx}=2 a x - \frac{b}{x^{2}}\)
    • Horizontal tangent at x = 1: \(2 a - b=0\)
    • The point is on the curve: \(a - 12 + b=0\)
    • a = 4, b = 8

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