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AI HL · AHL 2

Further Derivatives

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AHL 2.10

Mixed practice: stationary points, concavity and models

AHL 2.10

Mixed practice: stationary points, concavity and models

Target: I can sketch a graph from the signs of f′ and f″, and say what happens at a point of inflection. (+5 more)

By the end of this section

Key points

Reading f′ and f″

  • f′ > 0: the curve rises. f′ < 0: it falls. f′ = 0: stationary point.
  • f″ > 0: concave up (gradient rising). f″ < 0: concave down (gradient falling).
  • A point of inflection needs f″ = 0 and a change of sign. f″ = 0 alone is not enough.
  • On a graph of y = f′: the zeros of f′ are stationary points of f, and the maxima and minima of f′ are the points of inflection of f.
  • A stationary point of inflection is an inflection where f′ is also 0. On a graph of f′ the curve touches the x-axis there.

Finding and classifying

  • Write roots and reciprocals as powers first: \(\sqrt x=x^{1/2}\), \(1/x=x^{-1}\).
  • Solve f′ = 0 for x, then find y. Give exact values when the question asks for them.
  • Second derivative test: f″ < 0 is a maximum, f″ > 0 is a minimum. If f″ = 0, check the sign of f′ either side.
  • In a domain such as 0 < x < 2π, list every solution inside it and no others.

Constants and conditions

  • A minimum point (h, k) gives two equations: f′(h) = 0 and f(h) = k.
  • A curve concave up only for x > k means f″ = 0 at x = k.
  • A positive gradient at every point means f′ > 0 for all x. For a quadratic f′ that means no real roots: discriminant < 0.
  • Increasing for x < 0 and x > k means f′ = 0 at x = 0 and x = k.

Models and justification

  • Optimising: set the derivative to 0, then justify a maximum or minimum with f″ or with the sign of f′.
  • The greatest or least rate of change means optimising the derivative itself, so set the second derivative to 0.
  • A "show that" question gives the result: write every line of the working that leads to it.
  • Give the answer in the units of the model, and say what it means in the context.

On the calculator

Solving an equation numerically

  • Use it when an equation such as y″ = 2 cannot be solved by hand.
  • Type the equation and solve for x. Give the solution to 3 significant figures.
  • Check that the solution is inside the domain of the question.

Screenshot and video to come

Finding a minimum or maximum on a graph

  • Graph the function over a window that shows the turning points.
  • Use the minimum or maximum tool between two bounds that surround the point.
  • Use it to check your answer from calculus, not to replace the working.

Screenshot and video to come

Graphing f′ to read f

  • Graph the derivative. Where it crosses the x-axis, f has a stationary point.
  • Where the derivative has a maximum or a minimum, f has a point of inflection.
  • Check which side of zero the derivative is, to see whether f rises or falls.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Sketching from f′ and f″

Sketch a graph of \(y=f(x)\) where \(f'(x)>0\) for all x, \(f''(x)<0\) for \(x<2\) and \(f''(x)>0\) for \(x>2\). At which value of x does the curve change its bend?

Working
  1. Read f′

    \(f'>0\) everywhere

    The curve goes up as you move right, all the way. It never turns round.

  2. Read f″ for x < 2

    \(f''<0\): concave down

    The gradient is falling, so the curve bends downwards. It still rises, but more and more slowly.

  3. Read f″ for x > 2

    \(f''>0\): concave up

    The gradient is rising, so the curve bends upwards. It rises more and more steeply.

  4. −112345−551015xyy = f(x)
    The change

    Point of inflection at \(x=2\)

    f″ changes sign here, so the bend changes. The curve below is one example. Any increasing curve with this bend is a correct sketch.

Video walkthrough coming
A parameter from where the curve is concave up

The curve \(y=x^3+ax^2-5x+2\) is concave up for \(x>3\) only. (a) Find the value of a. (b) What happens to the curve at \(x=3\)?

Working
  1. Second derivative

    \(y'=3x^2+2ax-5,\quad y''=6x+2a\)

    Differentiate twice. Concave up means \(y''>0\).

  2. The boundary

    \(y''=0\) at \(x=3\): \(18+2a=0\)

    The curve is concave up only for x > 3, so y″ changes sign at x = 3. At that value y″ = 0.

  3. Solve

    \(a=-9\)

    Check: \(y''=6x-18\), which is positive only for x > 3.

  4. Part (b)

    Point of inflection at \(x=3\)

    y″ = 0 there and the sign changes from negative to positive. The curve changes from concave down to concave up.

Video walkthrough coming
Coefficients from a minimum point

The curve \(y=x^2+bx+c\) has a minimum point at \((3,-4)\). Find the values of b and c.

Working
  1. Gradient is zero

    \(y'=2x+b=0\) at \(x=3\): \(6+b=0\)

    A minimum is a stationary point, so the gradient is 0 there.

  2. Value of b

    \(b=-6\)

    This is the first unknown.

  3. The point is on the curve

    \(9+3(-6)+c=-4\)

    The minimum is the point (3, −4), so \(y=-4\) when \(x=3\).

  4. Solve

    \(c=5\)

    Check: \(y=x^2-6x+5\) has \(y(3)=-4\).

Video walkthrough coming
Stationary points of an exponential curve

Find the exact coordinates of the stationary points of \(y=e^{2x}-8e^x+6x\) and classify them. Give the y-coordinate of the minimum to 3 significant figures.

Working
  1. Differentiate

    \(y'=2e^{2x}-8e^x+6=2(e^x-1)(e^x-3)\)

    Treat \(e^{2x}\) as \((e^x)^2\) so the quadratic in \(e^x\) factorises.

  2. Stationary points

    \(e^x=1\) or \(e^x=3\), so \(x=0\) or \(x=\ln 3\)

    Set y′ = 0. Each factor gives one x.

  3. y-values

    \(y(0)=1-8=-7\), \(y(\ln 3)=9-24+6\ln 3=-15+6\ln 3\)

    Use \(e^{2\ln 3}=9\) and \(e^{\ln 3}=3\).

  4. Second derivative

    \(y''=4e^{2x}-8e^x\): \(y''(0)=-4\), \(y''(\ln 3)=12\)

    y″ < 0 gives a maximum. y″ > 0 gives a minimum.

  5. State

    Maximum \((0,-7)\); minimum \((\ln 3,\,-15+6\ln 3)\), where \(y=-8.41\)

    Check the second value with the calculator: \(-15+6\ln 3\).

Video walkthrough coming
Showing a curve is always concave down

Show that the graph of \(y=\ln(5-x)\) is concave down for every x in its domain. Find the value of \(y''\) at \(x=3\).

Working
  1. First derivative

    \(y'=\dfrac{-1}{5-x}\)

    Chain rule: the derivative of \(\ln u\) is \(u'/u\), and \(u'=-1\).

  2. Second derivative

    \(y'=-(5-x)^{-1}\Rightarrow y''=-(5-x)^{-2}=\dfrac{-1}{(5-x)^2}\)

    Differentiate \(-(5-x)^{-1}\) with the chain rule again. The two minus signs from \(-1\) and the inner derivative multiply to give a minus.

  3. Sign

    \((5-x)^2>0\), so \(y''<0\)

    A square is positive. The domain is \(x<5\), and y″ is never 0. So the curve is always concave down.

  4. The value

    \(y''(3)=\dfrac{-1}{2^2}=-0.25\)

    Put x = 3 into the second derivative.

Video walkthrough coming
Where a curve with ln x changes its bend

The curve is \(y=x^3\ln x\), for \(x>0\). Find the x-coordinate of its point of inflection, and give it to 3 significant figures.

Working
  1. First derivative

    \(y'=3x^2\ln x+x^3\cdot\dfrac1x=3x^2\ln x+x^2\)

    Product rule on \(x^3\ln x\).

  2. Second derivative

    \(y''=6x\ln x+3x^2\cdot\dfrac1x+2x=6x\ln x+5x=x(6\ln x+5)\)

    Product rule on the first term, ordinary rule on the second.

  3. Set y″ = 0

    \(x(6\ln x+5)=0\Rightarrow\ln x=-\dfrac56\)

    x > 0, so the factor x is not 0.

  4. Solve for x

    \(x=e^{-5/6}=0.435\)

    The sign of \(6\ln x+5\) changes here, so y″ changes sign. This is a point of inflection.

Video walkthrough coming
A quotient with ln x

The function f is defined by \(f(x)=\dfrac{\ln x}{x^2}\), for \(x>0\). (a) Find \(f'(x)\). (b) Find the exact coordinates of the stationary point. (c) Show that \(f''(x)=\dfrac{6\ln x-5}{x^4}\). (d) Hence classify the stationary point.

Working
  1. (a) Quotient rule

    \(f'=\dfrac{x^2\cdot\frac1x-2x\ln x}{x^4}=\dfrac{x-2x\ln x}{x^4}=\dfrac{1-2\ln x}{x^3}\)

    Divide top and bottom by x at the end.

  2. (b) Stationary point

    \(1-2\ln x=0\Rightarrow x=e^{1/2}\); \(f(e^{1/2})=\dfrac{1/2}{e}=\dfrac{1}{2e}\)

    A fraction is 0 when its numerator is 0. The point is \(\left(\sqrt e,\ \dfrac1{2e}\right)\).

  3. (c) Second derivative

    \(f''=\dfrac{-\frac2x\cdot x^3-3x^2(1-2\ln x)}{x^6}=\dfrac{-2x^2-3x^2+6x^2\ln x}{x^6}=\dfrac{6\ln x-5}{x^4}\)

    Quotient rule again, with \(u=1-2\ln x\) and \(v=x^3\). The result is the expression to show.

  4. (d) Classify

    \(f''(e^{1/2})=\dfrac{3-5}{e^2}=\dfrac{-2}{e^2}<0\)

    f″ is negative, so the stationary point is a local maximum.

Video walkthrough coming